Free math worksheets

- Find the greatest numerical factor shared by both coefficients. - For the variable part, use only factors of \(x\) that appear in both terms. - The greatest common monomial factor combines the common numerical and variable factors.

The greatest common numerical factor is \(6\), and both terms contain at least one factor of \(x\). Therefore the greatest common monomial factor is \(6x\).

\(6x\)

- First check whether the expression already has exactly two terms separated by subtraction. - Then check whether each entire term is a nonzero square of a polynomial with rational coefficients. - For a term such as \(cx^2\), ask whether \(c\) is the square of a rational number.

1. Choice B is \(x^2-4^2\), so it is directly a difference of two squares with rational coefficients. 2. Thus, \(P=x\) and \(Q=4\). 3. Choice A is a sum, and choice C has three terms as written. Choice D has two terms, but writing \(2x^2\) as a polynomial square would require the irrational coefficient \(\sqrt{2}\), so it does not meet the stated rational-coefficient condition.

B. \(x^2-16\), with \(P=x\) and \(Q=4\)

- Identify the numerical factor shared by both terms. - The expression inside the parentheses should reproduce both original terms when multiplied by \(4\).

1. Both terms share a factor of \(4\), so \(4x+12=4(x+3)\). 2. Distributing gives \(4(x+3)=4x+12\), which matches the original expression.

\(4(x+3)\). Check: \(4(x+3)=4x+12\).

- Find the greatest common factor of the numerical coefficients. - For each variable shared by all terms, use the smallest exponent present. - A term divided by itself leaves \(1\) inside the parentheses. - After dividing each term by the GCF, check that distributing restores the original expression.

1. In part a), the greatest common factor is \(6a^2b^2\), giving \(6a^2b^2(2a-3b+1)\). 2. In part b), the greatest common factor is \(12uv\), giving \(12uv(2u-3v)\). 3. In part c), the greatest common factor is \(4x\), giving \(4x(-x^2+3x-2)\). Stop here because the task asks only for greatest-common-factor extraction.

a) \(6a^2b^2(2a-3b+1)\) b) \(12uv(2u-3v)\) c) \(4x(-x^2+3x-2)\)

- Find the greatest common factor of the coefficients. - For each variable shared by all terms, use its smallest exponent. - When an entire term is factored out, a factor of \(1\) remains.

a) The greatest common factor is \(5u^2v^2\), so the expression factors as \(5u^2v^2(3u^2 + 2v)\). b) The greatest common factor is \(7x^3\), so the expression factors as \(7x^3(2x^2 - 3)\). c) The greatest common factor is \(3ab\), so the expression factors as \(3ab(3a - 2b + 1)\).

a) \(5u^2v^2(3u^2 + 2v)\) b) \(7x^3(2x^2 - 3)\) c) \(3ab(3a - 2b + 1)\)

- Multiply before adding or subtracting. - Simplify each product first. - Only terms with exactly the same variable part can be combined. - Track the sign of each product carefully.

1. For part a, multiply first: \((-3x)\cdot(-2y)=6xy\). Then \(6xy-8xy=-2xy\). 2. For part b, evaluate each product: \(4a\cdot(-b)=-4ab\) and \((-2a)\cdot(-3b)=6ab\). Then \(-4ab+6ab=2ab\). 3. For part c, the two factors of \(-1\) have product \(1\), so the first term is \(mn\). Therefore, \(mn-mn=0\).

a) \(-2xy\) b) \(2ab\) c) \(0\)

- For each numerator, look for a factor that matches the denominator. - Write the numerator as a product before canceling anything. - A factor may be canceled only when it is known to be nonzero.

1. Factor the first numerator: \(18a+18b=18(a+b)\). Then \(\frac{18(a+b)}{18}=a+b\). 2. Factor the second numerator: \(mp+np=p(m+n)\). Since \(p\ne0\), \(\frac{p(m+n)}{p}=m+n\). 3. Factor the third numerator: \(35x-21y=7(5x-3y)\). Then \(\frac{7(5x-3y)}{7}=5x-3y\). 4. The fourth numerator is already factored as \(12(u-v)\). Since \(u-v\ne0\), \(\frac{12(u-v)}{u-v}=12\).

1) Factored numerator: \(18(a+b)\); quotient: \(a+b\) 2) Factored numerator: \(p(m+n)\); quotient: \(m+n\) 3) Factored numerator: \(7(5x-3y)\); quotient: \(5x-3y\) 4) Factored numerator: \(12(u-v)\); quotient: \(12\)

- Find the greatest common factor of the coefficients. - Check whether every term contains a common variable. - Distribute mentally to check your result. - When factoring \(x\) from \(x^2\), one factor of \(x\) remains.

1) The greatest common factor is \(6\): \(12a + 18b = 6(2a + 3b)\). 2) The greatest common factor is \(x\): \(x^2 - 9x = x(x - 9)\). 3) The greatest common factor is \(5u\): \(15uv - 10uw = 5u(3v - 2w)\). 4) The greatest common factor is \(4k\): \(8k^2 + 12k = 4k(2k + 3)\).

1) \(6(2a + 3b)\) 2) \(x(x - 9)\) 3) \(5u(3v - 2w)\) 4) \(4k(2k + 3)\)

- Find the greatest common factor of the coefficients. - For each shared variable, use the smallest exponent. - When a whole variable factor is removed, a factor of \(1\) remains. - Distribute to check the factorization.

a) The greatest common factor is \(7xy\), so the expression factors as \(7xy(2x - 3y)\). b) The greatest common factor is \(3a^2\), so the expression factors as \(3a^2(a + 3)\). c) The greatest common factor is \(2km^2\), so the expression factors as \(2km^2(2k - 3m)\).

a) \(7xy(2x - 3y)\) b) \(3a^2(a + 3)\) c) \(2km^2(2k - 3m)\)

- Identify the factor present in every term. - For powers, use the smallest exponent shared by all terms. - Divide each term by the common factor to find what remains. - Distribute to check your answer.

1) The common factor is \(x\): \(x^2 + 8x = x(x + 8)\). 2) The lowest common power is \(c^2\): \(c^3 - c^2 = c^2(c - 1)\). 3) The common factor is \(a\): \(5a - ab = a(5 - b)\). 4) The lowest common power is \(y^3\): \(y^4 + y^3 = y^3(y + 1)\).

1) \(x(x + 8)\) 2) \(c^2(c - 1)\) 3) \(a(5 - b)\) 4) \(y^3(y + 1)\)

- Find the greatest common factor of the coefficients first. - Then identify the variables present in every term. - For each shared variable, use its smallest exponent. - The factored expression should have the same number of terms inside the parentheses as the original.

1) The greatest common factor is \(3a\): \(6ab - 9ac = 3a(2b - 3c)\). 2) The greatest common factor is \(5x^2\): \(10x^3 + 15x^2 = 5x^2(2x + 3)\). 3) The greatest common factor is \(4z\): \(4z^3 - 8z^2 + 12z = 4z(z^2 - 2z + 3)\). 4) The greatest common factor is \(uv\): \(u^2v + uv^2 = uv(u + v)\).

1) \(3a(2b - 3c)\) 2) \(5x^2(2x + 3)\) 3) \(4z(z^2 - 2z + 3)\) 4) \(uv(u + v)\)

- Find the greatest common factor of the coefficients. - Examine each variable separately to see whether it appears in every term. - For a shared variable, factor out its smallest exponent. - Distribute to verify the result.

a) The greatest common factor is \(7ab\), giving \(7ab(2a - 3b)\). b) The greatest common factor is \(5x\), giving \(5x(x^2 + 2x - 3)\). c) The greatest common factor is \(4uv^2\), giving \(4uv^2(2u + 3v)\).

a) \(7ab(2a - 3b)\) b) \(5x(x^2 + 2x - 3)\) c) \(4uv^2(2u + 3v)\)

- Find the greatest common factor of all coefficients. - Check which variables occur in every term. - For powers, use the smallest exponent common to all terms. - Remember that \(y = y(1)\). - Distribute to check the result.

1) The greatest common factor is \(4\): \(8a + 12b - 4c = 4(2a + 3b - c)\). 2) The greatest common factor is \(x^2\): \(x^3 - 5x^2 = x^2(x - 5)\). 3) The greatest common factor is \(6p\): \(6pq - 12pr + 6ps = 6p(q - 2r + s)\). 4) The greatest common factor is \(y\): \(y^2 + y = y(y + 1)\).

1) \(4(2a + 3b - c)\) 2) \(x^2(x - 5)\) 3) \(6p(q - 2r + s)\) 4) \(y(y + 1)\)

- Find the greatest positive numerical factor shared by all coefficients. - If an entire term is factored out, a factor of \(1\) or \(-1\) remains. - For a common variable, use the smallest exponent present in every term.

1) The greatest common factor is \(5xy\), so the expression factors as \(5xy(3x - 2y + 1)\). 2) The greatest common factor is \(3a\), so the expression factors as \(3a(-1 - 2b + 3c)\). 3) The greatest common factor is \(z^2\), so the expression factors as \(z^2(z^2 - z + 1)\).

1) \(5xy(3x - 2y + 1)\) 2) \(3a(-1 - 2b + 3c)\) 3) \(z^2(z^2 - z + 1)\)

- Identify the entire expression repeated in both terms. - Focus on what is inside each set of parentheses. - A term with no visible coefficient has a coefficient of \(1\). - You may temporarily treat the repeated binomial as one variable.

1) The common factor is \(x + y\), so the expression factors as \((6a + 5b)(x + y)\). 2) The common factor is \(p - 2\), so the expression factors as \((8m - 3)(p - 2)\). 3) Write the second term as \(1(v + 4)\). Factoring out \(v + 4\) gives \((u + 1)(v + 4)\).

1) \((6a + 5b)(x + y)\) 2) \((8m - 3)(p - 2)\) 3) \((u + 1)(v + 4)\)

- Identify the expression repeated in both terms. - Combine the factors outside the repeated parentheses into a new factor. - In part 3, the first repeated factor has an implied coefficient of \(1\). - Apply the distributive property in reverse.

1) Factor out \(y - z\): \((14x - 9w)(y - z)\). 2) Factor out \(m + 3\): \((k^2 + 7)(m + 3)\). 3) Write the first term as \(1(c - d)\). Factoring out \(c - d\) gives \((1 - 5a)(c - d)\).

1) \((14x - 9w)(y - z)\) 2) \((k^2 + 7)(m + 3)\) 3) \((1 - 5a)(c - d)\)

- Separate each expression into two large terms. - In part b, identify the implied coefficient before the second binomial. - The order of factors in a product does not change its value.

a) Factor out \(x + 2y\): \((4a - 5b)(x + 2y)\). b) The second term has coefficient \(-1\). Factoring out \(q - 1\) gives \((p - 1)(q - 1)\). c) Factor out \(u + v\): \((7 + w)(u + v)\).

a) \((4a - 5b)(x + 2y)\) b) \((p - 1)(q - 1)\) c) \((7 + w)(u + v)\)

- Find the greatest number dividing every coefficient. - Check which variables appear in every term. - If an entire term is removed, a factor of \(1\) remains. - Distribute to check the result.

1) The greatest common factor is \(6\): \(6(a - b + c)\). 2) The greatest common factor is \(xy\): \(xy(x + 1)\). 3) The greatest common factor is \(5z\): \(5z(4z^2 - 2z + 1)\). 4) The greatest common factor is \(4a\): \(4a(b - 2c + 3)\).

1) \(6(a - b + c)\) 2) \(xy(x + 1)\) 3) \(5z(4z^2 - 2z + 1)\) 4) \(4a(b - 2c + 3)\)

- Use the greatest positive common factor of the coefficients. - For fractions with the same denominator, compare their numerators. - Determine how many times the smaller decimal coefficient fits into the larger one. - For variables, factor out the smallest common exponent.

1) The greatest common factor is \(7uv\): \(7uv(2u - 3v + 1)\). 2) The greatest common factor is \(5x\): \(5x(-x^2 - 3x + 2)\). 3) The greatest common factor is \(0.6ab\): \(0.6ab(2a - 1)\). 4) The greatest common factor is \(\frac{2}{3}y\): \(\frac{2}{3}y(y + 2)\).

1) \(7uv(2u - 3v + 1)\) 2) \(5x(-x^2 - 3x + 2)\) 3) \(0.6ab(2a - 1)\) 4) \(\frac{2}{3}y(y + 2)\)

- Find the greatest common factor of the coefficients. - Identify variables present in every term. - For a shared variable, use the smallest exponent. - Distribute to check the result.

1) The greatest common factor is \(6a\): \(6a(2a + 3)\). 2) The greatest common factor is \(5xy\): \(5xy(3x - 2y)\). 3) The greatest common factor is \(3b\): \(3b(2b^2 - 3b + 1)\).

1) \(6a(2a + 3)\) 2) \(5xy(3x - 2y)\) 3) \(3b(2b^2 - 3b + 1)\)

- Find the greatest positive common factor of the coefficients. - For each variable, use the smallest exponent common to every term. - Track the signs of the remaining terms carefully. - When a whole term is divided by the common factor, a \(1\) or \(-1\) remains.

1) The greatest common factor is \(3rs\): \(3rs(3r - 4s)\). 2) The greatest common factor is \(7x^2y^2\): \(7x^2y^2(2x + 3y - 1)\). 3) The greatest common factor is \(5a^2b^3\): \(5a^2b^3(-2a^2 - 3ab + 5b^2)\).

1) \(3rs(3r - 4s)\) 2) \(7x^2y^2(2x + 3y - 1)\) 3) \(5a^2b^3(-2a^2 - 3ab + 5b^2)\)

- Identify the expression repeated in every term. - Determine what remains after removing the common expression. - A repeated expression standing alone has an implied coefficient of \(1\).

1) Factor out \(a - b\): \((4x + 3y - z)(a - b)\). 2) Factor out \(t^2 + 7\), including the implied coefficient \(1\) on the last term: \((s - 2r + 1)(t^2 + 7)\). 3) Factor out \(x + y - z\): \((u + v - w)(x + y - z)\).

1) \((4x + 3y - z)(a - b)\) 2) \((s - 2r + 1)(t^2 + 7)\) 3) \((u + v - w)(x + y - z)\)

- Look for an entire parenthetical expression repeated in every term. - Treat the repeated expression as one variable. - Preserve the operation between the outside factors.

a) Factor out \(x - y\): \((9a + 4b)(x - y)\). b) Factor out \(r + s - t\): \((u - v)(r + s - t)\).

a) \((9a + 4b)(x - y)\) b) \((u - v)(r + s - t)\)

- Write each term as a square. - Use \(A^2-B^2=(A-B)(A+B)\). - For a term such as \(9x^2\), take the square root of both the coefficient and the variable factor.

1) \(100-p^2=10^2-p^2=(10-p)(10+p)\). 2) \(9x^2-16=(3x)^2-4^2=(3x-4)(3x+4)\). 3) \(0.25-y^2=(0.5)^2-y^2=(0.5-y)(0.5+y)\). 4) \(m^2-\frac{4}{9}=m^2-\left(\frac{2}{3}\right)^2=\left(m-\frac{2}{3}\right)\left(m+\frac{2}{3}\right)\).

1) \((10-p)(10+p)\) 2) \((3x-4)(3x+4)\) 3) \((0.5-y)(0.5+y)\) 4) \(\left(m-\frac{2}{3}\right)\left(m+\frac{2}{3}\right)\)

- Express each term as a square. - Use \(A^2-B^2=(A-B)(A+B)\). - Keep decimal or fractional forms consistent within each factorization.

1) Since \(1.44m^2=(1.2m)^2\) and \(0.01n^2=(0.1n)^2\), the factorization is \((1.2m-0.1n)(1.2m+0.1n)\). 2) \(\frac{25}{36}-z^2=\left(\frac{5}{6}-z\right)\left(\frac{5}{6}+z\right)\). 3) \(225k^2-169h^2=(15k-13h)(15k+13h)\). 4) \(0.09p^2-\frac{1}{4}q^2=\left(0.3p-\frac{1}{2}q\right)\left(0.3p+\frac{1}{2}q\right)\).

1) \((1.2m-0.1n)(1.2m+0.1n)\) 2) \(\left(\frac{5}{6}-z\right)\left(\frac{5}{6}+z\right)\) 3) \((15k-13h)(15k+13h)\) 4) \(\left(0.3p-\frac{1}{2}q\right)\left(0.3p+\frac{1}{2}q\right)\)

- Rewrite each variable product as one squared expression. - Identify the square root of each numerical term. - Apply \(A^2-B^2=(A-B)(A+B)\).

1) \(r^2s^2-64=(rs)^2-8^2=(rs-8)(rs+8)\). 2) \(121-x^2y^2=11^2-(xy)^2=(11-xy)(11+xy)\). 3) \(a^2b^2-1=(ab)^2-1^2=(ab-1)(ab+1)\). 4) \(16-m^2n^2=4^2-(mn)^2=(4-mn)(4+mn)\).

1) \((rs-8)(rs+8)\) 2) \((11-xy)(11+xy)\) 3) \((ab-1)(ab+1)\) 4) \((4-mn)(4+mn)\)

- Find the square root of each numerical coefficient. - Treat a product such as \(abc\) as one factor before squaring. - Use the difference-of-squares identity.

1) \(0.25-x^2y^2=(0.5)^2-(xy)^2=(0.5-xy)(0.5+xy)\). 2) \(a^2b^2c^2-100=(abc)^2-10^2=(abc-10)(abc+10)\). 3) \(\frac{9}{16}-m^2n^2=\left(\frac{3}{4}\right)^2-(mn)^2=\left(\frac{3}{4}-mn\right)\left(\frac{3}{4}+mn\right)\). 4) \(0.01p^2q^2-49=(0.1pq)^2-7^2=(0.1pq-7)(0.1pq+7)\).

1) \((0.5-xy)(0.5+xy)\) 2) \((abc-10)(abc+10)\) 3) \(\left(\frac{3}{4}-mn\right)\left(\frac{3}{4}+mn\right)\) 4) \((0.1pq-7)(0.1pq+7)\)

- Check for subtraction between two square terms. - Identify the square root of each term. - A sum of squares does not match this identity.

a) \(81-x^2=9^2-x^2=(9-x)(9+x)\). b) This is a sum of squares, so the difference-of-squares identity does not apply. c) \(4a^2-121=(2a)^2-11^2=(2a-11)(2a+11)\). d) \(w^2-0.04=w^2-(0.2)^2=(w-0.2)(w+0.2)\). e) \(b^2-\frac{1}{9}=b^2-\left(\frac{1}{3}\right)^2=\left(b-\frac{1}{3}\right)\left(b+\frac{1}{3}\right)\).

a) \((9-x)(9+x)\) b) The identity does not apply because the expression is a sum. c) \((2a-11)(2a+11)\) d) \((w-0.2)(w+0.2)\) e) \(\left(b-\frac{1}{3}\right)\left(b+\frac{1}{3}\right)\)

- Treat each entire squared expression as one term. - Apply \(A^2-B^2=(A-B)(A+B)\). - Simplify inside each new factor.

1) \((x+5)^2-6^2=((x+5)-6)((x+5)+6)=(x-1)(x+11)\). 2) \(8^2-(a-3)^2=(8-(a-3))(8+(a-3))=(11-a)(a+5)\). 3) \((4y+1)^2-(3y)^2=((4y+1)-3y)((4y+1)+3y)=(y+1)(7y+1)\).

1) \((x-1)(x+11)\) 2) \((11-a)(a+5)\) 3) \((y+1)(7y+1)\)

- Treat the entire binomials as \(A\) and \(B\). - Distribute the minus sign in \((2x+3)-(x-1)\). - Combine like terms inside each factor.

1. Let \(A=2x+3\) and \(B=x-1\). 2. Then \(T=(A-B)(A+B)=((2x+3)-(x-1))((2x+3)+(x-1))\). 3. Simplifying the factors gives \((x+4)(3x+2)\).

\((x+4)(3x+2)\)

- Identify the two complete squared expressions. - Apply \(A^2-B^2=(A-B)(A+B)\). - Distribute subtraction signs before combining terms within each factor.

1) \((x+7)^2-5^2=(x+2)(x+12)\). 2) \((a-2b)^2-(3c+d)^2=(a-2b-3c-d)(a-2b+3c+d)\). 3) \((2k+1)^2-(k-3)^2=((2k+1)-(k-3))((2k+1)+(k-3))=(k+4)(3k-2)\).

1) \((x+2)(x+12)\) 2) \((a-2b-3c-d)(a-2b+3c+d)\) 3) \((k+4)(3k-2)\)

- Multiply each team's mean by its number of members to find its total points. - Divide the combined point total by the combined number of participants. - In part c), substitute \(m=n\), factor, and simplify.

a) Team A earned \(na\) total points, and Team B earned \(mb\) total points. Together, they earned \(na+mb\) points. b) There are \(n+m\) participants, so the combined mean is \(\frac{na+mb}{n+m}\). c) If \(n=m\), then \(\frac{na+nb}{n+n}=\frac{n(a+b)}{2n}=\frac{a+b}{2}\). This is the arithmetic mean of the two team means.

a) \(na+mb\) points b) \(\frac{na+mb}{n+m}\) points c) \(\frac{a+b}{2}\), the arithmetic mean of the two team means

- Form the difference between the two expressions. - Recognize the resulting trinomial as a binomial square. - Determine when that square equals \(0\).

1. Subtract the expressions: \(x^2+25-10x=x^2-10x+25\). 2. Rewrite the difference: \(x^2-10x+25=(x-5)^2\). 3. Since \((x-5)^2\ge0\), \(x^2+25\ge10x\) for every rational \(x\). 4. Equality occurs at \(x=5\), so the claim that the first expression is always strictly greater is false.

No. \(x^2+25-10x=(x-5)^2\ge0\), so \(x^2+25\ge10x\) for every rational \(x\). Equality occurs at \(x=5\), so the first expression is not always strictly greater.

- Find the greatest number dividing all coefficients. - Identify the variables present in every term. - Use the smallest common exponent for each shared variable. - Divide each term by the common factor. - A term equal to the common factor leaves \(1\).

1. The greatest common factor of the coefficients is \(5\). 2. Every term contains at least one factor of \(a\) and one factor of \(b\), so the greatest common factor is \(5ab\). 3. Divide each term by \(5ab\): \(3ab\), \(-2a^2\), \(4b^2\), and \(-1\). 4. The factored form is \(5ab(3ab - 2a^2 + 4b^2 - 1)\).

\(5ab(3ab - 2a^2 + 4b^2 - 1)\)

- Find the greatest common divisor of \(24\), \(16\), and \(32\). - Compare the exponents of \(x\), \(y\), and \(z\) across all terms. - Only variables present in every term can be factored out. - Distribute to check the result.

1. The greatest common factor of the coefficients is \(8\). 2. The smallest exponents common to all terms are \(x^3\), \(y^3\), and \(z\), so the greatest common factor is \(8x^3y^3z\). 3. Dividing each term gives \(3xz\), \(2y^2\), and \(-4z^2\). 4. The factored form is \(8x^3y^3z(3xz + 2y^2 - 4z^2)\).

\(8x^3y^3z(3xz + 2y^2 - 4z^2)\)

- Find the greatest number dividing both coefficients. - Use the smallest common exponent of each variable. - Determine what must remain inside the parentheses to reproduce each original term.

1. The greatest common factor of \(24\) and \(16\) is \(8\). 2. Both terms contain \(u^2v\), so the greatest common factor is \(8u^2v\). 3. Dividing the terms gives \(3v\) and \(-2u\). 4. The factored form is \(8u^2v(3v - 2u)\).

\(8u^2v(3v - 2u)\)

- Rewrite the larger power as a product containing \(y^n\). - Identify the factor common to both terms. - Distribute to check the result.

1. Rewrite \(y^{n+2}\) as \(y^ny^2\). 2. Both terms contain \(y^n\). 3. Factoring gives \(y^n(y^2 + 5)\).

\(y^n(y^2 + 5)\)

- Take the square root of each fractional coefficient. - Halve the even exponents in each squared monomial. - Use the difference-of-squares identity.

1. Rewrite the terms as squares: \(\frac{49}{81}a^4=\left(\frac{7}{9}a^2\right)^2\) and \(\frac{1}{16}b^2c^4=\left(\frac{1}{4}bc^2\right)^2\). 2. Apply the difference-of-squares identity: \(\left(\frac{7}{9}a^2-\frac{1}{4}bc^2\right)\left(\frac{7}{9}a^2+\frac{1}{4}bc^2\right)\).

\(\left(\frac{7}{9}a^2-\frac{1}{4}bc^2\right)\left(\frac{7}{9}a^2+\frac{1}{4}bc^2\right)\)

- Check for variables shared by every term, not only numerical factors. - If an entire term is factored out, a \(1\) or \(-1\) remains. - Use the greatest positive common factor of the coefficients. - Distribute to check your result.

1. The greatest common factor is \(4a\): \(4a(2x+3y)\). 2. The greatest common factor is \(7m\): \(7m(2-3n)\). 3. The greatest common factor is \(9\): \(9(-p-3q)\). 4. The greatest common factor is \(5u\): \(5u(3v+2w-1)\).

1) \(4a(2x+3y)\) 2) \(7m(2-3n)\) 3) \(9(-p-3q)\) 4) \(5u(3v+2w-1)\)

- Find the greatest common factor of the coefficients. - Identify the smallest common exponent of each variable. - Distribute to check the result. - Subtract exponents when dividing powers.

1) The greatest common factor is \(4x^2\): \(4x^2(x - 2)\). 2) The greatest common factor is \(6y^4\): \(6y^4(2y + 3)\). 3) The greatest common factor is \(5ab\): \(5ab(a - 3b)\). 4) The greatest common factor is \(8z^4\): \(8z^4(3z^3 - 2)\).

1) \(4x^2(x - 2)\) 2) \(6y^4(2y + 3)\) 3) \(5ab(a - 3b)\) 4) \(8z^4(3z^3 - 2)\)

- Compare \(x - 5\) and \(5 - x\). - After factoring the common binomial, inspect the coefficients in the new factor. - A complete factorization has no remaining common numerical factor.

1) Since \(-(5 - x) = x - 5\), the second term becomes \(+4b(x - 5)\). 2) Factor the common binomial: \((12a + 4b)(x - 5)\). 3) Factor \(4\) from \(12a + 4b\): \(4(3a + b)(x - 5)\).

1) \(+4b(x - 5)\) 2) \((12a + 4b)(x - 5)\) 3) \(4(3a + b)(x - 5)\)

- Find the identical parenthetical expression in every term. - Treat the common expression as one variable. - A parenthetical expression standing alone has an implied coefficient of \(1\).

a) Factor out \(a + b\): \((8x + 3y)(a + b)\). b) Factor out \(x - y\): \((p - q + r)(x - y)\). c) The second term has an implied coefficient of \(1\). Factoring out \(v^2 + 1\) gives \((u + 1)(v^2 + 1)\).

a) \((8x + 3y)(a + b)\) b) \((p - q + r)(x - y)\) c) \((u + 1)(v^2 + 1)\)

- Find each smaller rectangle's area from the side labels before adding them. - For the whole rectangle, combine the two horizontal pieces into one width. - Compare the two area expressions to see which quantity appears in every term of the sum.

1. The left rectangle has area \(hx\), and the right rectangle has area \(3h\). Their total area is \(hx+3h\). 2. The whole rectangle has height \(h\) and width \(x+3\), so its area is \(h(x+3)\). 3. The expressions are equivalent because both represent the same total area. Algebraically, factoring the common factor \(h\) from \(hx+3h\) gives \(h(x+3)\).

a) \(hx+3h\) b) \(h(x+3)\) c) They are equivalent representations of the same area; the common factor is \(h\).

- Look for a whole parenthesized expression that appears in both terms. - For the expanded form, distribute after combining the shared structure. - When choosing a form for \(x=4.05\), notice which expression makes the closeness of \(x\) to \(4\) immediately useful.

1. The two terms share the factor \((x-4)\), so \(E=(17+3)(x-4)=20(x-4)\). 2. Expanding gives \(E=20x-80\). 3. For \(x=4.05\), the factored form immediately gives \(E=20(4.05-4)=20(0.05)=1\). 4. The factored form is more efficient here because the small difference \(x-4\) is visible before any larger products are computed.

a) Factored: \(20(x-4)\); expanded: \(20x-80\) b) The factored form is more efficient, and \(E=1\).

- Before factoring something from a sum, check whether it is actually a factor of every term. - One way to test an alleged rewrite is to expand both expressions. - In the original expression, distribute first and then identify which terms are like terms.

1. The factor \((x+2)\) belongs only to the first term. The second term is \(5x\), not \(5(x+2)\), so the coefficients \(3\) and \(5\) cannot be combined as though both terms shared \((x+2)\). 2. Distributing correctly gives \(3(x+2)+5x=3x+6+5x\). 3. Combining like terms gives \(8x+6\). 4. Expanding Kai's expression gives \(8x+16\), confirming that it is not equivalent to \(8x+6\).

Kai treated \((x+2)\) as a common factor even though it is not a factor of \(5x\). The correct simplified form is \(8x+6\).

- Identify the number and variable shared by both terms. - Find the greatest common factor of \(6\) and \(9\). - Use the order of operations when evaluating. - Distribute to check the factored form.

1. The greatest common factor of \(6ax\) and \(9ay\) is \(3a\). 2. Therefore, \(6ax+9ay=3a(2x+3y)\). 3. The student is correct because distributing gives \(3a(2x)=6ax\) and \(3a(3y)=9ay\). 4. Original expression: \(6\cdot5\cdot2+9\cdot5\cdot4=60+180=240\). 5. Factored expression: \(3\cdot5(2\cdot2+3\cdot4)=15\cdot16=240\).

a) \(3a(2x+3y)\) b) Yes. Distributing \(3a\) reproduces the original expression. c) Both forms have the value \(240\).

- Check whether the coefficients inside each set of parentheses share a factor. - Check whether every term inside the parentheses still contains a common variable. - Find the greatest power of \(x\) common to both original terms.

1. The greatest common factor of \(60\) and \(45\) is \(15\). The lowest powers common to both terms are \(x^4\) and \(y^3\), so the greatest common factor is \(15x^4y^3\). 2. In A, the binomial \(12x - 9y\) still has a common factor of \(3\). 3. In B, the binomial \(4x^2 - 3xy\) still has a common factor of \(x\). 4. C factors out \(15x^4y^3\), and \(4x - 3y\) has no remaining common factor. Therefore, C is complete.

C is complete. In A, \(3\) can still be factored from the parentheses. In B, \(x\) can still be factored from the parentheses.

- Treat the repeated binomial as one object. - A minus sign before a binomial represents a coefficient of \(-1\). - Distribute to check each factored form.

a) The common factor is \(y - 2\), so the factored form is \((7x + 3)(y - 2)\). b) Write the second term as \(-1(b + 1)\). Factoring out \(b + 1\) gives \((5a - 1)(b + 1)\). c) The common factor is \(x + 3\), so the factored form is \((x^2 - 4)(x + 3)\).

a) \((7x + 3)(y - 2)\) b) \((5a - 1)(b + 1)\) c) \((x^2 - 4)(x + 3)\)

- Relate \(x - y\) and \(y - x\) by factoring out \(-1\). - For fractional coefficients, rewrite them with a common denominator. - For each common variable, use the smallest exponent present in every term.

a) Since \(y - x = -(x - y)\), the expression becomes \(a(x - y) - b(x - y)\). Factoring gives \((a - b)(x - y)\). b) The greatest common factor is \(9p^2q^2\). Factoring gives \(9p^2q^2(2p^2 - 3pq + 5q^2)\). c) The greatest common factor is \(\frac{1}{4}rs\). Factoring gives \(\frac{1}{4}rs(3rs - 2s + 1)\).

a) \((a - b)(x - y)\) b) \(9p^2q^2(2p^2 - 3pq + 5q^2)\) c) \(\frac{1}{4}rs(3rs - 2s + 1)\)

- Expand both products before comparing them. - Identify the terms that are identical. - Place the entire second expression in parentheses when subtracting.

a) \(T_1(x) = x^2 + 8x + 16\), and \(T_2(x) = x^2 + 8x + 12\). b) The quadratic and linear terms match, but \(16 > 12\), so \(T_1(x) > T_2(x)\) for every \(x\). c) \(T_1(x) - T_2(x) = 4\), a constant independent of \(x\).

a) \(T_1(x) = x^2 + 8x + 16\); \(T_2(x) = x^2 + 8x + 12\) b) \(T_1(x)\) is always greater. c) \(4\); the difference is constant.

- Find the greatest positive numerical factor shared by all coefficients. - With fractions that have the same denominator, compare their numerators. - Distribute to check your result.

a) The greatest common factor is \(0.5x^2y\), so the expression factors as \(0.5x^2y(-x + 3y)\). b) Factor out \(\frac{2}{7}a\): \(\frac{2}{7}a(a - 3)\). c) The greatest common factor is \(6k^2\), so the expression factors as \(6k^2(2k^2 - 3k + 1)\).

a) \(0.5x^2y(-x + 3y)\) b) \(\frac{2}{7}a(a - 3)\) c) \(6k^2(2k^2 - 3k + 1)\)

- Factor out the greatest common factor first. - Look for a difference of two squares in the remaining expression. - Continue until no factor can be factored further over the integers.

a) Factor out \(5\): \(5(x^2 - 9) = 5(x - 3)(x + 3)\). b) Factor out \(3ab\): \(3ab(a^2 - 4b^2) = 3ab(a - 2b)(a + 2b)\). c) Factor out \(y\): \(y(x^2 - 1) = y(x - 1)(x + 1)\).

a) \(5(x - 3)(x + 3)\) b) \(3ab(a - 2b)(a + 2b)\) c) \(y(x - 1)(x + 1)\)

- Treat \(x^n\) and \(y^m\) as distinct algebraic units. - Distribute each fractional coefficient carefully. - A minus sign before the second product changes the signs of its terms. - Combine only terms with the same base and exponent.

1. Expand each product: \(4x^n - 6y^m - 6x^n + 9y^m + 2x^n + 3y^m\). 2. Combine the \(x^n\)-terms: \((4 - 6 + 2)x^n = 0\). 3. Combine the \(y^m\)-terms: \((-6 + 9 + 3)y^m = 6y^m\). 4. The result is \(6y^m\).

\(6y^m\)

- Expand each product separately. - What happens to the terms containing \(x\) when you subtract the expressions? - If the simplified difference contains no variable, its value is the same for every input.

a) Expand: \(A = x^2 + 7x + 10\) and \(B = x^2 + 7x + 12\). Then \(B - A = (x^2 + 7x + 12) - (x^2 + 7x + 10) = 2\), so the difference is constant. b) \(B-A=2\). c) Since the difference is constant, \(B-A=2\) when \(x=-3.5\).

a) \(A = x^2 + 7x + 10\), \(B = x^2 + 7x + 12\), and \(B - A = 2\) b) \(2\) c) \(2\)

- Simplify each expression independently. - Expand the binomial squares in \(T_1\). - Distribute the factor in \(T_2\), then compare the simplified results.

1. Expand \(T_1\): \((5x+2)^2+(2x-5)^2=(25x^2+20x+4)+(4x^2-20x+25)\). 2. Combine like terms: \(T_1(x)=29x^2+29\). 3. Distribute in \(T_2\): \(T_2(x)=29x^2+29\). 4. Because both expressions simplify to \(29x^2+29\), they are equivalent for every real \(x\).

Both expressions simplify to \(29x^2+29\), so \(T_1(x)=T_2(x)\) for every real \(x\).

- First look for a greatest common factor in each expanded polynomial. - Check whether the remaining expression is a difference of squares or a perfect-square trinomial. - You can verify a match by expanding the proposed factored form.

1. For \(f_1(x) = x^3 - 25x\), factor out \(x\): \(x(x^2 - 25)\). Then factor the difference of squares: \(x(x - 5)(x + 5)\). Thus, \(f_1\) matches \(g_c\). 2. For \(f_2(x) = x^3 + 10x^2 + 25x\), factor out \(x\): \(x(x^2 + 10x + 25) = x(x + 5)^2\). Thus, \(f_2\) matches \(g_a\). 3. For \(f_3(x) = 3x^2 - 12x + 12\), factor out \(3\): \(3(x^2 - 4x + 4) = 3(x - 2)^2\). Thus, \(f_3\) matches \(g_b\).

\(f_1 \to g_c\); \(f_2 \to g_a\); \(f_3 \to g_b\)

- Represent consecutive integers as \(n\) and \(n + 1\). - Translate the words “product” and “subtract” into operations. - Write one expression for the entire statement before simplifying. - Expand the product and combine like terms.

1. Let \(n\) be the smaller integer. The next integer is \(n + 1\). 2. The described expression is \(n(n + 1) - n\). 3. Expand and simplify: \(n(n + 1) - n = n^2 + n - n = n^2\). 4. Therefore, the result is always the square of the smaller integer.

Let the smaller integer be \(n\). Then \(n(n + 1) - n = n^2\), so the statement is true for every integer \(n\).

- Represent the middle integer with a variable. - Express the integers immediately before and after it. - Look for a product of a sum and a difference.

1. Let the middle integer be \(m\). Then the three integers are \(m-1\), \(m\), and \(m+1\). 2. The product of the outer integers, plus \(1\), is \((m-1)(m+1)+1\). 3. Use the difference-of-squares identity: \((m-1)(m+1)+1=m^2-1+1=m^2\). 4. This is exactly the square of the middle integer.

Let the integers be \(m-1\), \(m\), and \(m+1\). Then \((m-1)(m+1)+1=m^2-1+1=m^2\), which is the square of the middle integer.

- Represent the two consecutive integers with a variable. - Expand the binomial square. - Rewrite the result in the form \(2k+1\), where \(k\) is an integer.

1. Let the consecutive integers be \(n\) and \(n+1\), where \(n\) is an integer. 2. Their squared sum is \(n^2+(n+1)^2\). 3. Expand and combine: \(n^2+n^2+2n+1=2n^2+2n+1=2(n^2+n)+1\). 4. Since \(n^2+n\) is an integer, \(2(n^2+n)\) is even. Adding \(1\) makes the result odd.

\(n^2+(n+1)^2=2(n^2+n)+1\). This has the form \(2k+1\) for an integer \(k=n^2+n\), so it is always odd.

- Factor the variable shared by both terms in the numerator. - What happens when a nonzero factor appears in both the numerator and denominator? - Test your conclusion with simple values of \(x\), \(y\), and \(z\).

a) Factor the numerator: \(xz + yz = z(x + y)\). Since \(x + y \ne 0\), cancel the common factor: \(z(x + y) \div (x + y) = z\). b) When \(z = 42\), the value is \(42\). c) The factor \(x + y\) cancels completely, so only \(z\) remains. Therefore, the value does not depend on \(x\) and \(y\), provided \(x + y \ne 0\).

a) \(z\) b) \(42\) c) The common factor \(x + y\) cancels when \(x + y \ne 0\), leaving only \(z\).

- Work backward by multiplying the desired quotient by the denominator. - Compare corresponding coefficients after expanding. - Determine exactly when the denominator of the claimed quotient equals zero.

1. For the first identity, the numerator must equal \(13(x+y)=13x+13y\), so \(r=13\) and \(s=13\). 2. For the second identity, \(5(3z-4)=15z-20\), so \(c=20\). 3. Since \(ax+ay=a(x+y)\), the quotient equals \(a\) only when \(x+y\ne0\). If \(x+y=0\), equivalently \(x=-y\), the denominator is zero and the quotient is undefined.

1. a) \(r=13\), \(s=13\) b) \(c=20\) 2. The claim is valid only when \(x+y\ne0\). The quotient is undefined when \(x=-y\).

- Divide each original term by the factor being removed. - When a whole term is factored out, a factor of \(1\) remains. - For the reverse task, compare one term of the product with the corresponding term in the expanded expression. - Preserve the signs of the original terms.

1. Dividing the terms by \(7x\) gives \(2y\) and \(-3z\), so \(14xy-21xz=7x(2y-3z)\). 2. Dividing by \(ab\) gives \(a\) and \(b\), so \(a^2b+ab^2=ab(a+b)\). 3. The outside factor is \(12m^2\div4m=3m\), and \(3m\cdot(-5)=-15m\), so the monomial is \(3m\). 4. Dividing each term by \(3\) gives \(2p\), \(-3q\), and \(1\), so \(6p-9q+3=3(2p-3q+1)\).

1) \(7x(2y-3z)\) 2) \(ab(a+b)\) 3) \(3m\) 4) \(3(2p-3q+1)\)

- Divide each original term by the factor being removed. - Inspect the terms inside the parentheses for another common factor. - Compare the factors outside the parentheses. - A greatest-common-factor form leaves no common factor inside.

a) Factoring out \(6xy\) gives \(6xy(4x + 6y)\). b) The greatest common factor of the coefficients is \(12\), and both terms contain \(xy\). Thus, the greatest common factor is \(12xy\), giving \(12xy(2x + 3y)\). c) In part a), the terms inside the parentheses still share a factor of \(2\). In part b), \(2x\) and \(3y\) have no common factor other than \(1\).

a) \(6xy(4x + 6y)\) b) \(12xy(2x + 3y)\) c) The first set of parentheses still has a common factor of \(2\); the second does not.

- Distribute the outside factor to each term inside the parentheses. - Inspect the terms inside each set of parentheses for a remaining common factor. - A complete greatest-common-factor form leaves no common factor inside. - Compare the two outside factors.

a) Tim: \(8u^2v^2(3u) = 24u^3v^2\) and \(8u^2v^2(2v) = 16u^2v^3\), so his expression matches the original. Mia: \(4uv(6u^2v) = 24u^3v^2\) and \(4uv(4uv^2) = 16u^2v^3\), so her expression also matches the original. b) In Mia's parentheses, both terms still share \(2uv\). In Tim's parentheses, \(3u\) and \(2v\) have no common factor other than \(1\). Therefore, Tim factored out the greatest common factor \(8u^2v^2\), while Mia factored out only part of it.

a) Both factored expressions expand to \(24u^3v^2 + 16u^2v^3\). b) Tim removed the greatest common factor. Mia's parentheses still contain a common factor of \(2uv\).

- A greatest-common-factor form leaves no common factor inside the parentheses. - Compare the variables in both students' parentheses. - Distribute the outside factor to check the expression.

a) Ava factored out \(4x\), while Ben factored out \(4xy^2\). In Ava's parentheses, both terms still contain \(y^2\), so her common factor is not the greatest one. In Ben's parentheses, \(3x\) and \(2y\) have no common numerical or variable factor other than \(1\). Therefore, Ben factored out the greatest common factor. b) \(4xy^2(3x - 2y) = 12x^2y^2 - 8xy^3\).

a) Ben factored out the greatest common factor. Ava's parentheses still contain a common factor of \(y^2\). b) \(4xy^2(3x - 2y) = 12x^2y^2 - 8xy^3\)

- Find the greatest positive common factor of the coefficients first. - For each variable, use the smallest exponent found in every term. - When a whole term is removed, a factor of \(1\) or \(-1\) remains. - Distribute to check every sign in the result.

1) The greatest common factor is \(6ab\), giving \(6ab(2a - 3b)\). 2) The greatest common factor is \(5x^2y^2\), giving \(5x^2y^2(3x + 2y - 1)\). 3) The greatest common factor is \(4u^2v^2\), giving \(4u^2v^2(-2u^2 - 3uv + v^2)\).

1) \(6ab(2a - 3b)\) 2) \(5x^2y^2(3x + 2y - 1)\) 3) \(4u^2v^2(-2u^2 - 3uv + v^2)\)

- Identify factors present in every term. - Use the greatest positive common factor of the coefficients. - Distribute to check the result. - With several variables, verify each variable separately.

1) The greatest common factor is \(7p^2q^2\), so the expression factors as \(7p^2q^2(3p^2q - 2pq^2 + 1)\). 2) The greatest common factor is \(8x^2y^2\), so the expression factors as \(8x^2y^2(-2x^3 - 3xy^2 - 1)\). 3) The greatest common factor is \(6rst\), so the expression factors as \(6rst(2t + 3r - 5s)\).

1) \(7p^2q^2(3p^2q - 2pq^2 + 1)\) 2) \(8x^2y^2(-2x^3 - 3xy^2 - 1)\) 3) \(6rst(2t + 3r - 5s)\)

- Factoring out a negative value reverses every sign inside the parentheses. - Work systematically: coefficients first, then each variable. - Subtract exponents when dividing variable powers. - Check that no common factor remains inside the parentheses.

1) The greatest common factor of the coefficients is \(15\), and the lowest common powers are \(u^2v^2w\). The result is \(15u^2v^2w(3u - 2v + 4w)\). 2) The greatest common factor of the coefficient magnitudes is \(9\), and the lowest common powers are \(x^2y^2\). Factoring out \(-9x^2y^2\) gives \(-9x^2y^2(2x^2 - 3xy + y^2)\).

1) \(15u^2v^2w(3u - 2v + 4w)\) 2) \(-9x^2y^2(2x^2 - 3xy + y^2)\)

- Factoring out \(-1\) reverses the order in a difference. - Compare the binomials and determine how they are related. - An expression with no visible coefficient has a coefficient of \(1\). - Rewrite the second binomial so it matches the first.

1) Since \(v - u = -(u - v)\), the expression becomes \(3(u - v) - w(u - v) = (3 - w)(u - v)\). 2) Since \(4 - x = -(x - 4)\), the expression becomes \(7(x - 4) + y(x - 4) = (7 + y)(x - 4)\). 3) Since \(c - b = -(b - c)\), the expression becomes \(a(b - c) + (b - c) = (a + 1)(b - c)\).

1) \((3 - w)(u - v)\) 2) \((7 + y)(x - 4)\) 3) \((a + 1)(b - c)\)

- Use the identity \(b - a = -(a - b)\). - Transform the left side of part 3 one step at a time. - Look for a common binomial in every part. - In part 2, account for the two negative signs.

1) Since \(y - x = -(x - y)\), the expression becomes \(2k(x - y) - 5m(x - y) = (2k - 5m)(x - y)\). 2) Since \(b - a = -(a - b)\), the expression becomes \(x^2(a - b) + 4(a - b) = (x^2 + 4)(a - b)\). 3) Since \(q - p = -(p - q)\), the left side becomes \(z(p - q) - (p - q)\). Factoring out \(p - q\) gives \((z - 1)(p - q)\), which is the right side.

1) \((2k - 5m)(x - y)\) 2) \((x^2 + 4)(a - b)\) 3) Rewriting \(q - p\) as \(-(p - q)\) makes \(p - q\) a common factor, producing the right side.

- Use the least exponent of each common variable factor. - Factor numerical greatest common factors as well. - Distribute your factored form mentally to check it.

1) The greatest common factor is \(5z^n\), giving \(5z^n(3z^2 + 2)\). 2) Factor out \(x^a\): \(x^a(x^b - 1)\). 3) Factor out \(2^k\): \(2^k(2^2 - 1) = 3 \cdot 2^k\). 4) Factor out \(r^ns^n\): \(r^ns^n(s^n - r^n)\).

1) \(5z^n(3z^2 + 2)\) 2) \(x^a(x^b - 1)\) 3) \(3 \cdot 2^k\) 4) \(r^ns^n(s^n - r^n)\)

- Simplify the expression before substituting values. - Evaluate the factor \(c - d\) for the given values. - Any product containing a factor of zero equals zero.

1) Since \(d - c = -(c - d)\), \(T = 5x(c - d) + 4y(c - d) = (5x + 4y)(c - d)\). 2) Substitute: \(T = (5 \cdot 2 + 4 \cdot 3)(8 - 8) = 22 \cdot 0 = 0\). 3) When \(c = d\), the factor \(c - d\) is zero. Therefore, the product is zero for all values of \(x\) and \(y\).

1) \((5x + 4y)(c - d)\) 2) \(0\) 3) When \(c = d\), the value is always \(0\), regardless of \(x\) and \(y\).

- After factoring once, check whether any resulting factor can be factored further. - Look for a numerical factor in a new binomial. - Reversing a subtraction inside parentheses introduces a negative sign.

1) Since \(5 - v = -(v - 5)\), the expression becomes \(2u(v - 5) - 4(v - 5) = (2u - 4)(v - 5)\). Factor \(2\) from the first factor to get \(2(u - 2)(v - 5)\). 2) Since \(3 - y = -(y - 3)\), the expression becomes \(y^2(y - 3) + 9(y - 3) = (y^2 + 9)(y - 3)\).

1) \(2(u - 2)(v - 5)\) 2) \((y^2 + 9)(y - 3)\)

- Compare the two binomials in each expression. - Reversing the order of a subtraction changes its sign. - A binomial standing alone has an implied coefficient of \(1\).

a) Since \(2 - x = -(x - 2)\), the expression becomes \(5a(x - 2) - 3(x - 2) = (5a - 3)(x - 2)\). b) Since \(q - p = -(p - q)\), the expression becomes \((p - q) + 4y(p - q) = (1 + 4y)(p - q)\). c) Since \(b - a = -(a - b)\), the expression becomes \(0.5z(a - b) - (a - b) = (0.5z - 1)(a - b)\).

a) \((5a - 3)(x - 2)\) b) \((1 + 4y)(p - q)\) c) \((0.5z - 1)(a - b)\)

- Determine how the minus sign affects \(y-x\). - Rewrite both terms using exactly the same binomial. - Count the total number of factors of \(x-y\).

1. The differences \(x-y\) and \(y-x\) are opposites, so \(-(y-x)=x-y\). 2. Rewrite the expression: \(4(x-y)-(y-x)=4(x-y)+(x-y)\). 3. Combine the common factors: \((4+1)(x-y)=5(x-y)\). 4. Therefore, Jordan is correct.

Jordan is correct: \(4(x-y)-(y-x)=4(x-y)+(x-y)=(4+1)(x-y)=5(x-y)\).

- Use the smaller exponent on each common base. - A term divided by itself leaves \(1\). - Check by distributing the common factor.

1) Factor out \(x^n\): \(x^n(1 + x^2)\). 2) Factor out \(3y^k\): \(3y^k(2y - 1)\). 3) Factor out \(a^m\): \(a^m(a^m - 1)\). 4) Factor out \(x^ny^n\): \(x^ny^n(x + y)\).

1) \(x^n(1 + x^2)\) 2) \(3y^k(2y - 1)\) 3) \(a^m(a^m - 1)\) 4) \(x^ny^n(x + y)\)

- Distribute the student’s expression and compare it with the original. - Count the terms in the original expression and inside the parentheses. - When a term equals the common factor, dividing it by that factor gives \(1\).

1. Distribute the student’s expression: \(8p^2q^2(3p - 2q) = 24p^3q^2 - 16p^2q^3\). 2. The original expression has a third term, \(+8p^2q^2\), which is missing from the student’s result. 3. Dividing that third term by the common factor gives \(8p^2q^2 \div 8p^2q^2 = 1\). 4. The correct factorization is \(8p^2q^2(3p - 2q + 1)\).

The factorization is incorrect because it omits the third term. The correct result is \(8p^2q^2(3p - 2q + 1)\).

- Compare the signs and order inside the binomials. - Rewrite \(-y - 1\) as the negative of one binomial. - Use \(a - b = -(b - a)\) to create matching factors.

a) Factor out \(y + z\): \((2x - 5)(y + z)\). b) Since \(4 - b = -(b - 4)\), the expression becomes \(a(b - 4) - 3(b - 4) = (a - 3)(b - 4)\). c) Group the last two terms as \(-(y + 1)\): \(x(y + 1) - (y + 1) = (x - 1)(y + 1)\). d) Since \(w - v = -(v - w)\), the expression becomes \(4u(v - w) + 7t(v - w) = (4u + 7t)(v - w)\).

a) \((2x - 5)(y + z)\) b) \((a - 3)(b - 4)\) c) \((x - 1)(y + 1)\) d) \((4u + 7t)(v - w)\)

- Use the quotient rule for exponents when dividing variable powers. - Distribute to check the expression in part a. - Examine the coefficients and variables left inside the parentheses after factoring out \(3mn\).

a) Divide each term by \(6m^2n^2\): \(12m^5n^3 \div 6m^2n^2 = 2m^3n\), \(-18m^3n^4 \div 6m^2n^2 = -3mn^2\), and \(30m^2n^2 \div 6m^2n^2 = 5\). Therefore, the expression inside the parentheses is \(2m^3n - 3mn^2 + 5\). b) Factoring out only \(3mn\) leaves \(4m^4n^2 - 6m^2n^3 + 10mn\), whose terms still share \(2mn\). Thus, \(3mn\) is not the greatest common factor.

a) \(2m^3n - 3mn^2 + 5\) b) The remaining terms still have a common factor of \(2mn\), so the factorization is incomplete.

- Compare the order of the variables inside the binomials. - A binomial with no visible coefficient has a coefficient of \(1\). - After factoring once, check whether a numerical factor remains in another factor.

1) Since \(z - y = -(y - z)\), the expression becomes \(7x(y - z) - 2(y - z) = (7x - 2)(y - z)\). 2) The second binomial has an implied coefficient of \(-1\), so factoring gives \((5a - 1)(b + c)\). 3) First factor out \(t - 1\): \((6s + 3)(t - 1)\). Then factor \(3\) from \(6s + 3\), giving \(3(2s + 1)(t - 1)\).

1) \((7x - 2)(y - z)\) 2) \((5a - 1)(b + c)\) 3) \(3(2s + 1)(t - 1)\)

- Group the terms without parentheses so they match the existing binomial. - When placing a minus sign before a new set of parentheses, reverse the signs inside. - If a whole binomial is removed, an implied factor of \(1\) remains.

a) Rewrite \(-y + 7\) as \(-(y - 7)\). Then \(x(y - 7) - (y - 7) = (x - 1)(y - 7)\). b) Rewrite \(3 - c\) as \(-(c - 3)\). Then \(2a(c - 3) - (c - 3) = (2a - 1)(c - 3)\).

a) \((x - 1)(y - 7)\) b) \((2a - 1)(c - 3)\)

- For each expression, look for factors shared by every term. - In part b), treat the fractional coefficient and the variable factors separately when checking what is common. - In part c), compare one term of the expanded product \(M(2u - 3v)\) with the matching term on the left.

1. For part a), the greatest common factor is \(8x^2y^2\). Dividing each term by it gives \(3x^2\), \(-2xy\), and \(1\), so the factorization is \(8x^2y^2(3x^2 - 2xy + 1)\). 2. For part b), dividing each term by \(\frac{2}{5}ab\) gives \(a\) and \(-2b\). Thus \(\frac{2}{5}a^2b - \frac{4}{5}ab^2 = \frac{2}{5}ab(a - 2b)\). 3. For part c), comparing \(M(2u - 3v)\) with the original expression gives \(M = 6uvw\), since \(6uvw(2u - 3v) = 12u^2vw - 18uv^2w\).

a) \(8x^2y^2(3x^2 - 2xy + 1)\) b) \(\frac{2}{5}ab(a - 2b)\) c) \(M = 6uvw\)

- Combine terms that contain the same binomial. - Reversing the order of a difference changes its sign. - Rewrite the expression so every repeated binomial is identical before factoring.

a) Since \(t - s = -(s - t)\), the middle term becomes \(+4r(s - t)\). Then \((s - t) + 4r(s - t) + 2(s - t) = (3 + 4r)(s - t)\). b) Since \(2y - x = -(x - 2y)\), the expression becomes \(5a(x - 2y) - (x - 2y) = (5a - 1)(x - 2y)\).

a) \((3 + 4r)(s - t)\) b) \((5a - 1)(x - 2y)\)

- Reverse a subtraction by factoring out \(-1\). - Use \(b - a = -(a - b)\). - A binomial with no visible coefficient has a coefficient of \(1\).

1) Since \(3 - x = -(x - 3)\), the expression becomes \(a(x - 3) - b(x - 3) = (a - b)(x - 3)\). 2) Since \(-(n - m) = m - n\), the expression becomes \(5(m - n) + k(m - n) = (5 + k)(m - n)\). 3) The second term has an implied coefficient of \(-1\), so factoring gives \((x - 1)(p + q)\).

1) \((a - b)(x - 3)\) 2) \((5 + k)(m - n)\) 3) \((x - 1)(p + q)\)

- Opposite binomials differ by a factor of \(-1\). - Group unparenthesized terms so they match the existing binomial. - When placing a minus before parentheses, reverse all signs inside.

1) Since \(y - x = -(x - y)\), the expression becomes \(5(x - y) - a(x - y) = (5 - a)(x - y)\). 2) Group the last two terms as \(-(q + 1)\): \(p(q + 1) - (q + 1) = (p - 1)(q + 1)\). 3) Group the last two terms as \(-(v - 3)\): \(2u(v - 3) - (v - 3) = (2u - 1)(v - 3)\).

1) \((5 - a)(x - y)\) 2) \((p - 1)(q + 1)\) 3) \((2u - 1)(v - 3)\)

- Identify the implied coefficient before a binomial standing alone. - Look for a structure that can be made common to every term. - Temporarily treating a binomial as one object may make the pattern clearer. - Reverse a subtraction by factoring out \(-1\).

1) Since \(-(v - u) = u - v\), the expression becomes \(z(u - v) + (u - v) = (z + 1)(u - v)\). 2) Since \(3 - 2x = -(2x - 3)\), the expression becomes \(5a(2x - 3) - 4b(2x - 3) = (5a - 4b)(2x - 3)\). 3) Since \(b - a = -(a - b)\), the expression becomes \(x(a - b) - y(a - b) + (a - b) = (x - y + 1)(a - b)\).

1) \((z + 1)(u - v)\) 2) \((5a - 4b)(2x - 3)\) 3) \((x - y + 1)(a - b)\)

- First factor within part of the expression to create a repeated binomial. - After creating the common binomial, factor it from the entire expression. - In the final item, check whether another numerical common factor remains after the first factoring step.

1. Factor \(5\) from the last two terms: \(a(x+y)+5(x+y)\). Then factor out \(x+y\): \((a+5)(x+y)\). 2. Factor \(c\) from the last two terms: \(b(m-n)+c(m-n)\). Then factor out \(m-n\): \((b+c)(m-n)\). 3. Factor \(6\) from the last two terms: \(2x(a+b)+6(a+b)\). Factor out \(a+b\): \((2x+6)(a+b)\). Then factor \(2\) from the first factor to obtain \(2(x+3)(a+b)\).

1) \((a+5)(x+y)\) 2) \((b+c)(m-n)\) 3) \(2(x+3)(a+b)\)

- Reversing a difference introduces a factor of \(-1\). - If no common factor appears across the whole expression, group pairs of terms. - In part 3, write \(y^2\) as \(y \cdot y\).

1) Factor \(3\) from the last two terms: \(k(x - 2) + 3(x - 2)\). Then factor out \(x - 2\): \((k + 3)(x - 2)\). 2) Since \(b - a = -(a - b)\), the expression becomes \(p(a - b) - q(a - b) = (p - q)(a - b)\). 3) Group terms: \((y^2 + 4y) + (xy + 4x) = y(y + 4) + x(y + 4)\). Factor out \(y + 4\): \((x + y)(y + 4)\).

1) \((k + 3)(x - 2)\) 2) \((p - q)(a - b)\) 3) \((x + y)(y + 4)\)

- Place parentheses around the ungrouped terms so they match the existing binomial. - A binomial standing alone has an implied coefficient of \(1\). - Pay attention to signs when grouping terms after a minus sign.

1) Group the last two terms as \(1(a + b)\): \(4s(a + b) + (a + b) = (4s + 1)(a + b)\). 2) Group the last two terms as \(-(y - z)\): \(3x(y - z) - (y - z) = (3x - 1)(y - z)\). 3) Group the last two terms as \(-(m + n)\): \(k(m + n) - (m + n) = (k - 1)(m + n)\).

1) \((4s + 1)(a + b)\) 2) \((3x - 1)(y - z)\) 3) \((k - 1)(m + n)\)

- Factor from the unparenthesized terms so their remaining expression matches the existing binomial. - A negative factor reverses all signs inside parentheses. - Reorder a difference by factoring out \(-1\).

1) Factor \(3\) from the last two terms: \(5x(a - 2) + 3(a - 2) = (5x + 3)(a - 2)\). 2) Factor \(-3\) from the last two terms: \(2y(c + d) - 3(c + d) = (2y - 3)(c + d)\). 3) Since \(v - u = -(u - v)\), the expression becomes \(z(u - v) - (u - v) = (z - 1)(u - v)\).

1) \((5x + 3)(a - 2)\) 2) \((2y - 3)(c + d)\) 3) \((z - 1)(u - v)\)

- Track signs carefully when creating groups. - For the second method, pair terms differently from the first method. - Look for a common factor in \(12xy\) and \(-3y\). - Distribute to check the final product.

a) Group as \((12xy - 8x) + (-3y + 2)\). Factor each group: \(4x(3y - 2) - (3y - 2)\). Then factor the common binomial: \((4x - 1)(3y - 2)\). b) Alternatively, group as \((12xy - 3y) + (-8x + 2)\). Then \(3y(4x - 1) - 2(4x - 1) = (3y - 2)(4x - 1)\), the same product because multiplication is commutative.

a) \((4x - 1)(3y - 2)\) b) \((12xy - 3y) + (-8x + 2) = 3y(4x - 1) - 2(4x - 1) = (3y - 2)(4x - 1)\)

- Divide the four terms into two pairs with a common factor in each pair. - Factor each pair as much as possible. - Then look for a whole binomial common to both new terms. - Preserve the subtraction in part c.

a) Group and factor: \(5(x + y) + a(x + y) = (5 + a)(x + y)\). b) Group and factor: \(z(z + 3) + m(z + 3) = (z + m)(z + 3)\). c) Group and factor: \(7(r - s) + k(r - s) = (7 + k)(r - s)\).

a) \((5 + a)(x + y)\) b) \((z + m)(z + 3)\) c) \((7 + k)(r - s)\)

- Reorder the terms before grouping if useful. - The second factoring step requires the same binomial in both groups. - Use the distributive property to determine the needed constant.

a) Group as \((ab + 4a) + (3b + 12)\). Then \(a(b + 4) + 3(b + 4) = (a + 3)(b + 4)\). b) Regroup as \((ab + 3b) + (4a + 12)\). Then \(b(a + 3) + 4(a + 3) = (b + 4)(a + 3)\), the same product in reverse order. c) The second group must be \(3(b + 4)\), whose constant term is \(3 \cdot 4 = 12\). A different constant would not create the same binomial factor.

a) \((a + 3)(b + 4)\) b) \((ab + 3b) + (4a + 12) = b(a + 3) + 4(a + 3) = (b + 4)(a + 3)\) c) The constant must equal \(3 \cdot 4 = 12\) so both groups contain the same binomial factor.

- Look for a grouping that creates the same binomial in two places. - After factoring, substitute into the simpler product rather than the original four-term expression. - Check that expanding your factored form would recover all four original terms.

1. Group the terms: \((6x^2-12x)+(5x-10)\). 2. Factor each group: \(6x(x-2)+5(x-2)\). 3. Factor out the common binomial: \((6x+5)(x-2)\). 4. Substitute \(x=12\): \((72+5)(10)=77\cdot10=770\).

Factored form: \((6x+5)(x-2)\) Value: \(770\)

- Track the sign when factoring the second group. - A product is zero when either factor is zero. - Use the structure of the factored form in part c.

a) Group as \((4ab - 8a) + (-3b + 6)\). Factor each group: \(4a(b - 2) - 3(b - 2)\). Then factor the common binomial: \((4a - 3)(b - 2)\). b) Substitute: \((4 \cdot 1.5 - 3)(5 - 2) = (6 - 3) \cdot 3 = 9\). c) If \(b = 2\), then \(b - 2 = 0\). Therefore, the product equals \(0\) for every value of \(a\).

a) \((4a - 3)(b - 2)\) b) \(9\) c) \(0\)

- Pair terms that share a common factor. - Factor each pair as much as possible. - Look for the same binomial after the first factoring step. - The final result should be a product of two binomials.

1) Group the terms: \((4am - 6bm) + (2an - 3bn)\). Factor each group: \(2m(2a - 3b) + n(2a - 3b)\). Then factor the common binomial: \((2a - 3b)(2m + n)\). 2) Group the terms: \((x^2 + 5x) + (xy + 5y)\). Factor each group: \(x(x + 5) + y(x + 5)\). Then factor the common binomial: \((x + 5)(x + y)\).

1) \((2a - 3b)(2m + n)\) 2) \((x + 5)(x + y)\)

- Group the terms in pairs. - Track signs carefully when factoring a negative value from a group. - Make the binomial remaining from both groups identical. - Then factor out that common binomial.

1) Group as \((12ac - 8bc) + (-15ad + 10bd)\). Factor each group: \(4c(3a - 2b) - 5d(3a - 2b)\). Therefore, the factorization is \((3a - 2b)(4c - 5d)\). 2) Group as \((21x^2 - 14xy) + (-6x + 4y)\). Factor each group: \(7x(3x - 2y) - 2(3x - 2y)\). Therefore, the factorization is \((3x - 2y)(7x - 2)\).

1) \((3a - 2b)(4c - 5d)\) 2) \((3x - 2y)(7x - 2)\)

- Track the minus signs inside each group. - Factor the greatest common factor from the first pair. - Factor the second pair so the same binomial appears. - Distribute to check the final result.

1) Group and factor: \((3ab - 6a) + (2b - 4) = 3a(b - 2) + 2(b - 2)\). Therefore, the result is \((3a + 2)(b - 2)\). 2) Group and factor: \((x^3 - 4x^2) + (3x - 12) = x^2(x - 4) + 3(x - 4)\). Therefore, the result is \((x^2 + 3)(x - 4)\).

1) \((3a + 2)(b - 2)\) 2) \((x^2 + 3)(x - 4)\)

- Divide the expression into two pairs. - Factor the first pair. - Factor a negative value from the second pair so both binomials match. - A binomial standing alone has an implied coefficient of \(1\).

1. Group the terms: \((5am - 5an) + (-m + n)\). 2. Factor each group: \(5a(m - n) - (m - n)\). 3. Factor the common binomial: \((5a - 1)(m - n)\).

\((5a - 1)(m - n)\)

- Separate the first three terms from the last three terms. - Look for the same sum of variables in both groups. - Factor the common trinomial. - Then factor the remaining expression in \(x\).

1. Group by powers of \(x\): \((ux^2 + vx^2 + wx^2) + (ux + vx + wx)\). 2. Factor each group: \(x^2(u + v + w) + x(u + v + w)\). 3. Factor the common trinomial: \((x^2 + x)(u + v + w)\). 4. Factor \(x\) from \(x^2 + x\): \(x(x + 1)(u + v + w)\).

\(x(x + 1)(u + v + w)\)

- Track minus signs carefully while factoring each group. - Find the greatest common factor in each pair. - Factoring out a negative value reverses the signs inside parentheses. - Distribute to verify each product.

1) Group and factor: \((xy - 3x) + (2y - 6) = x(y - 3) + 2(y - 3)\). Therefore, \((x + 2)(y - 3)\). 2) Group and factor: \((a^2 - ab) + (-7a + 7b) = a(a - b) - 7(a - b)\). Therefore, \((a - 7)(a - b)\). 3) Group and factor: \((4km + 2kn) + (-6m - 3n) = 2k(2m + n) - 3(2m + n)\). Therefore, \((2k - 3)(2m + n)\).

1) \((x + 2)(y - 3)\) 2) \((a - 7)(a - b)\) 3) \((2k - 3)(2m + n)\)

- Look for the same sum appearing after a simple grouping step. - Treat a repeated parenthesized expression as one algebraic factor. - Verify the factorization before substituting the numerical values.

1. Group the terms: \((2x^2+6x)+(5xy+15y)\). 2. Factor each group: \(2x(x+3)+5y(x+3)\). 3. Factor out the common binomial: \((2x+5y)(x+3)\). 4. Substitute \(x=2\) and \(y=1.5\): \((4+7.5)(5)=11.5\cdot5=57.5\).

Factored form: \((2x+5y)(x+3)\) Value: \(57.5\)

- Group the expression into two pairs and compare what remains after factoring each pair. - Keep track of the negative coefficient in the second group. - Check the factorization before substituting the decimal values.

1. Group the terms: \((6a^2+9ab)+(-4ac-6bc)\). 2. Factor each group: \(3a(2a+3b)-2c(2a+3b)\). 3. Factor out the common binomial: \((3a-2c)(2a+3b)\). 4. Substitute the values: \((3.6-1)(2.4+1.2)=2.6\cdot3.6=9.36\).

Factored form: \((3a-2c)(2a+3b)\) Value: \(9.36\)

- Check whether the operation is subtraction. - Determine whether both terms are squares over the rational numbers. - Apply \(A^2-B^2=(A-B)(A+B)\) only when both conditions hold.

a) Since \(121=11^2\), \(a^2-121=(a-11)(a+11)\). b) This is a sum of squares, not a difference of squares, so the identity does not apply. c) Since \(16x^2=(4x)^2\), \(16x^2-1=(4x-1)(4x+1)\). d) Factoring as \((x-r)(x+r)\) would require a rational number \(r\) with \(r^2=20\). No such rational number exists, so it does not factor this way over the rational numbers.

a) \((a-11)(a+11)\) b) It cannot be factored by the difference-of-squares identity because it is a sum. c) \((4x-1)(4x+1)\) d) It does not factor as a difference of rational squares because \(20\) is not the square of a rational number.

- Remember that \(x^4=(x^2)^2\). - Check the operation between the two square terms. - For part c), replace the difference of squares with a product.

1. Since \(0.04x^4=(0.2x^2)^2\) and \(\frac{1}{25}y^2=\left(\frac{1}{5}y\right)^2\), part a) factors as \(\left(0.2x^2-\frac{1}{5}y\right)\left(0.2x^2+\frac{1}{5}y\right)\). 2. The identity \(A^2-B^2=(A-B)(A+B)\) requires subtraction, so it does not apply to the sum \(81a^2+16b^2\). 3. For part c), \(51^2-49^2=(51-49)(51+49)=2\cdot100=200\).

a) \(\left(0.2x^2-\frac{1}{5}y\right)\left(0.2x^2+\frac{1}{5}y\right)\) b) It is a sum of squares, not a difference of squares. c) \(51^2-49^2=(51-49)(51+49)=200\)

- Use each entire binomial as \(A\) or \(B\). - Distribute a minus sign across every term inside parentheses. - Combine like terms within each resulting factor.

1) \(((a+b)-(a-b))((a+b)+(a-b))=(2b)(2a)=4ab\). 2) \(((2x+3y)-(x+y))((2x+3y)+(x+y))=(x+2y)(3x+4y)\). 3) Since \(0.49k^2=(0.7k)^2\), the factors are \((0.7k-(0.3k-1))(0.7k+(0.3k-1))=(0.4k+1)(k-1)\).

1) \(4ab\) 2) \((x+2y)(3x+4y)\) 3) \((0.4k+1)(k-1)\)

- Rewrite the nonsquared term as a square when needed. - Treat an expression inside parentheses as one complete term. - Simplify all nested parentheses after factoring.

1) Since \(9a^2=(3a)^2\), the factorization is \((3a-(3b-c))(3a+(3b-c))=(3a-3b+c)(3a+3b-c)\). 2) Since \(16z^2=(4z)^2\), the factorization is \((x+2y-4z)(x+2y+4z)\). 3) Since \(0.49=(0.7)^2\), the factorization is \((0.7-(u+v))(0.7+(u+v))=(0.7-u-v)(0.7+u+v)\).

1) \((3a-3b+c)(3a+3b-c)\) 2) \((x+2y-4z)(x+2y+4z)\) 3) \((0.7-u-v)(0.7+u+v)\)

- Rewrite each first term as a square when necessary. - Treat each parenthesized expression as one complete term. - After applying the identity, check whether numerical factors can still be factored out.

1) Since \(x^2y^4=(xy^2)^2\), the factorization is \((xy^2-z-2)(xy^2+z+2)\). 2) Since \(0.01=(0.1)^2\), the factorization is \((0.1-(x-0.1))(0.1+(x-0.1))=(0.2-x)x\). 3) Write \(9m^2=(3m)^2\). Then \((3m-(m+4))(3m+(m+4))=(2m-4)(4m+4)=8(m-2)(m+1)\).

1) \((xy^2-z-2)(xy^2+z+2)\) 2) \(x(0.2-x)\) 3) \(8(m-2)(m+1)\)

- Factor the difference of squares instead of expanding both squares. - Simplify the difference and sum factors separately. - Show the final expression as \(24\) times an integer.

1. Factor: \(T=((2n+3)-(2n-3))((2n+3)+(2n-3))\). 2. Simplify the factors: \(T=6\cdot4n=24n\). 3. Since \(n\) is a natural number, \(24n\) is a multiple of \(24\).

\(T=((2n+3)-(2n-3))((2n+3)+(2n-3))=6\cdot4n=24n\). Since \(n\) is a natural number, \(T\) is always a multiple of \(24\).

- Subtract \(4ab\) from \((a+b)^2\). - Expand and rewrite the result as a binomial square. - Determine when that square equals \(0\).

1. Form the difference: \((a+b)^2-4ab\). 2. Expand and simplify: \(a^2+2ab+b^2-4ab=a^2-2ab+b^2=(a-b)^2\). 3. Since \((a-b)^2\ge0\), it follows that \((a+b)^2\ge4ab\). 4. Equality occurs exactly when \((a-b)^2=0\), which means \(a=b\).

\((a+b)^2-4ab=(a-b)^2\ge0\), so \((a+b)^2\ge4ab\). Equality occurs exactly when \(a=b\).

- Use the labeled pieces of an outer side to determine the full side length of the large square. - Compare the area of the whole square with the combined area of the four rectangles. - For part c), identify which part of the large square is not included in \(A_2\), then use \(a>b\) when interpreting a squared side length.

1. Each side of the large square is \(a+b\), so \(A_1=(a+b)^2\). Each of the four rectangles has area \(ab\), so \(A_2=4ab\). 2. Subtracting gives \(A_1-A_2=(a+b)^2-4ab=a^2+2ab+b^2-4ab=a^2-2ab+b^2=(a-b)^2\). 3. The four rectangles and the uncovered interior region fill the large square, so \((a-b)^2\) is the area of the interior region. That region is a square. Since \(a>b\), its positive side length is \(a-b\).

a) \(A_1=(a+b)^2\) and \(A_2=4ab\) b) \(A_1-A_2=(a-b)^2\) c) \((a-b)^2\) is the area of the uncovered interior square, whose side length is \(a-b\).

- First view the expression as one square minus another. - Check whether either resulting factor is also a difference of squares. - Stop when every factor is irreducible over the rational numbers.

1. Write the expression as a difference of squares: \((n^2)^2-1^2=(n^2-1)(n^2+1)\). 2. Factor \(n^2-1\) again: \(n^2-1=(n-1)(n+1)\). 3. The polynomial \(n^2+1\) does not factor over the rational numbers. 4. Therefore, \(n^4-1=(n-1)(n+1)(n^2+1)\), which has three nonconstant factors.

\(n^4-1=(n-1)(n+1)(n^2+1)\); there are three nonconstant factors.

- Look for a structure that is shared by whole terms before trying to factor smaller pieces. - A factor that remains after one step may itself have a useful algebraic structure. - In the last expression, treat the entire binomial inside the square as one algebraic quantity.

1. The first expression is a difference of squares: \(36x^2-49y^2=(6x)^2-(7y)^2\). 2. Therefore it factors as \((6x-7y)(6x+7y)\). 3. In the second expression, first factor out the common binomial: \(4a^2(b+c)-9(b+c)=(b+c)(4a^2-9)\). The remaining factor is a difference of squares, so the complete factorization is \((b+c)(2a-3)(2a+3)\). 4. For the third expression, treat \(z+2\) as one quantity: \((z+2)^2-5^2\). 5. Apply the difference-of-squares pattern to obtain \((z+2-5)(z+2+5)=(z-3)(z+7)\).

a) \((6x-7y)(6x+7y)\) b) \((b+c)(2a-3)(2a+3)\) c) \((z-3)(z+7)\)

- Group the terms in pairs. - Factor the common factor from each pair. - Compare the binomials that remain. - Factor out the common binomial.

a) Group and factor: \((ax + 3a) + (bx + 3b) = a(x + 3) + b(x + 3)\). Therefore, \((a + b)(x + 3)\). b) Group and factor: \((y^2 - 5y) + (zy - 5z) = y(y - 5) + z(y - 5)\). Therefore, \((y + z)(y - 5)\).

a) \((a + b)(x + 3)\) b) \((y + z)(y - 5)\)

- Expand the student’s factors to check them. - Find the expression whose square is \(16a^2\). - Apply the difference-of-squares identity.

1. The student halved \(16\) instead of finding its square root. Since \((8a)^2=64a^2\), the proposed factors do not reproduce the original expression. 2. The correct square terms are \(16a^2=(4a)^2\) and \(81=9^2\). 3. Therefore, \(16a^2-81=(4a-9)(4a+9)\).

The coefficient should be square-rooted, not divided by \(2\). The correct factorization is \((4a-9)(4a+9)\).

- Take square roots of fractional or decimal coefficients. - Halve even exponents when identifying a squared monomial. - Apply \(A^2-B^2=(A-B)(A+B)\).

1) \(\frac{1}{16}u^{12}-v^8=\left(\frac{1}{4}u^6\right)^2-(v^4)^2=\left(\frac{1}{4}u^6-v^4\right)\left(\frac{1}{4}u^6+v^4\right)\). 2) \(1.21m^2n^{10}-0.04k^4=(1.1mn^5)^2-(0.2k^2)^2=(1.1mn^5-0.2k^2)(1.1mn^5+0.2k^2)\). 3) \(36a^2b^6-\frac{25}{81}c^4=(6ab^3)^2-\left(\frac{5}{9}c^2\right)^2=\left(6ab^3-\frac{5}{9}c^2\right)\left(6ab^3+\frac{5}{9}c^2\right)\).

1) \(\left(\frac{1}{4}u^6-v^4\right)\left(\frac{1}{4}u^6+v^4\right)\) 2) \((1.1mn^5-0.2k^2)(1.1mn^5+0.2k^2)\) 3) \(\left(6ab^3-\frac{5}{9}c^2\right)\left(6ab^3+\frac{5}{9}c^2\right)\)

- Check the operation between the square terms. - Rewrite \(x^{10}\) as \((x^5)^2\). - Identify decimal coefficients that are perfect squares.

1) \(1.21m^2-0.04n^2=(1.1m)^2-(0.2n)^2=(1.1m-0.2n)(1.1m+0.2n)\). 2) \(36x^{10}-1=(6x^5)^2-1^2=(6x^5-1)(6x^5+1)\). 3) The expression is a sum of squares, not a difference of squares, so this identity does not apply over the rational numbers.

1) \((1.1m-0.2n)(1.1m+0.2n)\) 2) \((6x^5-1)(6x^5+1)\) 3) The difference-of-squares identity does not apply because the expression is a sum.

- Before choosing a factoring pattern, check whether all terms share a numerical, variable, or parenthesized factor. - Treat a repeated parenthesized expression as a single algebraic unit. - After one factoring step, inspect the remaining factors to see whether any can be factored further.

1. Factor out \(2\) from the first expression: \(2x^2-18y^2=2(x^2-9y^2)\). 2. The remaining difference of squares factors as \(2(x-3y)(x+3y)\). 3. In the second expression, factor out the repeated binomial \((b-c)\) to get \((b-c)(6a+9d)\), then factor out \(3\). This gives \(3(b-c)(2a+3d)\). 4. In the third expression, treat \((x+1)\) as one quantity and use a difference of squares: \(z^2-(x+1)^2=(z-(x+1))(z+(x+1))\). 5. Simplifying the binomial factors gives \((z-x-1)(z+x+1)\).

1) \(2(x-3y)(x+3y)\) 2) \(3(b-c)(2a+3d)\) 3) \((z-x-1)(z+x+1)\)

- Look for three terms that form a binomial square. - Group terms to expose a common factor. - After one factoring step, check for a difference of squares.

1) Rewrite the first three terms as \((x+3)^2\): \((x+3)^2-(2y)^2=(x+3-2y)(x+3+2y)\). 2) Rewrite the last three terms as \((a+b)^2\): \(49-(a+b)^2=(7-a-b)(7+a+b)\). 3) Group: \(a(x+y)+(x+y)^2\). Factor out \(x+y\): \((x+y)(a+x+y)\).

1) \((x+3-2y)(x+3+2y)\) 2) \((7-a-b)(7+a+b)\) 3) \((x+y)(a+x+y)\)

- Recognize the numerator as a difference of squares. - Use \(x^{\frac{2}{3}}=\left(x^{\frac{1}{3}}\right)^2\). - Factor before canceling. - Interpret an exponent of \(\frac{1}{3}\) as a cube root.

a) Write the numerator as a difference of squares: \(x^{\frac{2}{3}}-y^{\frac{2}{3}}=\left(x^{\frac{1}{3}}\right)^2-\left(y^{\frac{1}{3}}\right)^2\). Therefore \(x^{\frac{2}{3}}-y^{\frac{2}{3}}=\left(x^{\frac{1}{3}}-y^{\frac{1}{3}}\right)\left(x^{\frac{1}{3}}+y^{\frac{1}{3}}\right)\). Since the denominator is positive, cancel the common factor to obtain \(E=x^{\frac{1}{3}}-y^{\frac{1}{3}}\). b) For \(x=27\) and \(y=8\), \(E=27^{\frac{1}{3}}-8^{\frac{1}{3}}=3-2=1\).

a) \(E=x^{\frac{1}{3}}-y^{\frac{1}{3}}\) b) \(1\)

- When the outside factor is given, divide each original term by it. - When the inside factor is given, divide a corresponding original term by it. - Track signs and variable exponents carefully.

1) Divide by \(7xy\): the terms inside are \(2x\) and \(-3y\), so the blank is \(2x - 3y\). 2) The outside factor is \(9a^4 \div 3a^2 = 3a^2\). It also gives \(6a^2 \div 2 = 3a^2\). 3) Divide by \(5b^3\): the terms inside are \(4b^2\) and \(-5\), so the blank is \(4b^2 - 5\). 4) Divide by \(6u^2v^2\): the terms inside are \(2u\) and \(3v\), so the blank is \(2u + 3v\).

1) \(2x - 3y\) 2) \(3a^2\) 3) \(4b^2 - 5\) 4) \(2u + 3v\)

- Reversing the order of a difference introduces a factor of \(-1\). - Rewrite one binomial so it matches the other exactly. - Use \(b - a = -(a - b)\).

1) Since \(b - a = -(a - b)\), the expression becomes \(x(a - b) - y(a - b) = (x - y)(a - b)\). 2) Since \(t - 2 = -(2 - t)\), the expression becomes \(s^2(2 - t) + 5(2 - t) = (s^2 + 5)(2 - t)\). 3) Since \(n - m = -(m - n)\), the expression becomes \(7(m - n) - k(m - n) = (7 - k)(m - n)\).

1) \((x - y)(a - b)\) 2) \((s^2 + 5)(2 - t)\) 3) \((7 - k)(m - n)\)

- Look for a common numerical factor in addition to the common binomial. - When an entire term is factored out, a factor of \(-1\) may remain. - Distribute mentally to check the result.

a) Factor out \(x - 5\): \((12a - 18b)(x - 5)\). Then factor \(6\) from the first factor: \(6(2a - 3b)(x - 5)\). b) Every term contains \(5(p + q)\). Factoring it out gives \(5(x + y - 1)(p + q)\).

a) \(6(2a - 3b)(x - 5)\) b) \(5(x + y - 1)(p + q)\)

- Group the terms in pairs. - Pay attention to signs when the third term is negative. - Distribute to check each factored expression.

1) Group and factor: \((2x^2 - 6xy) + (-5x + 15y) = 2x(x - 3y) - 5(x - 3y)\). Thus, \((2x - 5)(x - 3y)\). 2) Group and factor: \((12ab - 9ac) + (8b - 6c) = 3a(4b - 3c) + 2(4b - 3c)\). Thus, \((3a + 2)(4b - 3c)\). 3) Group and factor: \((14p^2 - 7pq) + (-10p + 5q) = 7p(2p - q) - 5(2p - q)\). Thus, \((7p - 5)(2p - q)\).

1) \((2x - 5)(x - 3y)\) 2) \((3a + 2)(4b - 3c)\) 3) \((7p - 5)(2p - q)\)

- Track the negative sign on the third term. - Factoring out a negative value reverses all signs inside the parentheses. - Make the binomial factors identical before the final factoring step.

1) Group and factor: \((x^2 - 5x) + (-xy + 5y) = x(x - 5) - y(x - 5)\). Therefore, \((x - y)(x - 5)\). 2) Group and factor: \((14uv - 21u) + (-10v + 15) = 7u(2v - 3) - 5(2v - 3)\). Therefore, \((7u - 5)(2v - 3)\).

1) \((x - y)(x - 5)\) 2) \((7u - 5)(2v - 3)\)

- Compare the two products before expanding them and look for an entire factor that repeats. - After extracting the repeated structure, simplify what remains inside the other factor. - To prove divisibility by \(6\), aim for a form that is \(6\) times an integer.

1. Both products contain the factor \((m-3)\), so the expression can be rewritten as \((m-3)((m+2)-(m-4))\). 2. The expression inside the second factor simplifies to \(6\), giving \(6(m-3)\). 3. Since \(m\) is an integer, \(m-3\) is an integer. Therefore \(6(m-3)\) is divisible by \(6\).

\(6(m-3)\). Since \(m-3\) is an integer, the expression is always divisible by \(6\).

- In part a), focus on the repeated parenthesized quantity rather than expanding it. - Ask what happens to any term containing \(b+c\) when that quantity equals \(0\). - For part c), distribute each outside coefficient carefully before combining anything.

1. The first two terms share the factor \((b+c)\), so \(P=(4a-7)(b+c)+3\). 2. When \(b+c=0\), the product term is \(0\), so \(P=3\). 3. Expanding the original expression gives \(P=4ab+4ac-7b-7c+3\). 4. The partly factored form makes the effect of \(b+c=0\) immediate, while the expanded form displays the individual terms and their coefficients.

a) \(P=(4a-7)(b+c)+3\) b) \(P=3\) c) \(P=4ab+4ac-7b-7c+3\) d) The partly factored form exposes what happens when \(b+c=0\); the expanded form exposes the individual terms and coefficients.

- Identify the factor repeated in both terms on the left. - Factoring first may be more efficient than expanding every product. - Simplify the expression inside the brackets before solving. - Substitute your solution into the original equation to check it.

1. Factor out \(x - 3\): \((x - 3)[4(x + 5) - (4x + 2)] = 36\). 2. Simplify the bracket: \(4x + 20 - 4x - 2 = 18\). 3. Solve \(18(x - 3) = 36\): \(x - 3 = 2\), so \(x = 5\). 4. Check: \(4 \cdot 2 \cdot 10 - 22 \cdot 2 = 80 - 44 = 36\).

\(x = 5\)

- Reorder the terms if the first grouping does not produce matching binomials. - Track signs when factoring a negative value from a group. - The two groups must leave exactly the same binomial factor.

a) Reorder and group: \((12ac + 8bc) + (-15ad - 10bd)\). Factor: \(4c(3a + 2b) - 5d(3a + 2b)\). Therefore, \((3a + 2b)(4c - 5d)\). b) Reorder and group: \((x^2 - xy) + (7x - 7y)\). Factor: \(x(x - y) + 7(x - y)\). Therefore, \((x + 7)(x - y)\). c) Group: \((18ab - 27ac) + (-10b + 15c)\). Factor: \(9a(2b - 3c) - 5(2b - 3c)\). Therefore, \((9a - 5)(2b - 3c)\).

a) \((3a + 2b)(4c - 5d)\) b) \((x + 7)(x - y)\) c) \((9a - 5)(2b - 3c)\)

- A binomial identity may be applied more than once. - Treat the larger parenthesized expressions as single terms. - For part 2, compare the structure with \(A^2+2AB+B^2\).

1) Treat the expression as a difference of squares: \(((x^2+9)-6x)((x^2+9)+6x)\). Each factor is a perfect-square trinomial, so \((x^2-6x+9)(x^2+6x+9)=(x-3)^2(x+3)^2=(x^2-9)^2\). 2) Let \(A=x+y\) and \(B=x-y\). Then \(A^2+2AB+B^2=(A+B)^2\). Since \((x+y)+(x-y)=2x\), the result is \((2x)^2=4x^2\).

1) \((x-3)^2(x+3)^2\), equivalently \((x^2-9)^2\) 2) \(4x^2\)

- Before choosing a binomial identity, check whether all terms share a greatest common factor. - In part 2, look for a three-term expression that can be treated as one square. - In part 3, try grouping terms so the same binomial factor appears twice. - After each factoring step, inspect the remaining factors for another usable structure.

1. First factor out the GCF: \(3y^3-12y=3y(y^2-4)\). Then factor the difference of squares: \(3y(y-2)(y+2)\). 2. Rewrite the first three terms as \((a-b)^2\): \((a-b)^2-4^2=(a-b-4)(a-b+4)\). 3. Group the terms: \(x^2(x+3)-4(x+3)=(x+3)(x^2-4)=(x+3)(x-2)(x+2)\).

1) \(3y(y-2)(y+2)\) 2) \((a-b-4)(a-b+4)\) 3) \((x+3)(x-2)(x+2)\)

- Look for the greatest power of \(n\) common to every term. - Group the terms in pairs to reveal a common binomial factor. - Check for a difference of squares after grouping. - A fifth-degree expression can contain five linear factors when multiplicity is counted.

1. Factor out the greatest common factor: \(P(n) = n^2(n^3 + n^2 - 4n - 4)\). 2. Factor the cubic by grouping: \(n^3 + n^2 - 4n - 4 = n^2(n + 1) - 4(n + 1)\). 3. Factor out the common binomial: \((n + 1)(n^2 - 4)\). 4. Factor the difference of squares: \(n^2 - 4 = (n - 2)(n + 2)\). 5. Since \(n^2=n\cdot n\), the complete linear-factor form is \(P(n)=n\cdot n\cdot(n+1)(n-2)(n+2)\). Thus, \(n\) is a linear factor with multiplicity \(2\).

\(n\cdot n\cdot(n+1)(n-2)(n+2)\); the factor \(n\) has multiplicity \(2\).

- Rewrite \(x-y\) as a difference of cubes. - Use the factorization \(a^3-b^3=(a-b)(a^2+ab+b^2)\). - Simplify the expression in parentheses before multiplying by the outside factor. - Add exponents when multiplying powers with the same base.

1. View the numerator as a difference of cubes: \(x-y=(x^{\frac{1}{3}})^3-(y^{\frac{1}{3}})^3\). 2. Factor: \(x-y=(x^{\frac{1}{3}}-y^{\frac{1}{3}})(x^{\frac{2}{3}}+x^{\frac{1}{3}}y^{\frac{1}{3}}+y^{\frac{2}{3}})\). 3. Cancel the common factor in the fraction to obtain \(x^{\frac{1}{3}}-y^{\frac{1}{3}}\). 4. The expression in parentheses becomes \(x^{\frac{1}{3}}-y^{\frac{1}{3}}+y^{\frac{1}{3}}=x^{\frac{1}{3}}\). 5. Therefore \(B=x^{\frac{1}{3}}x^{-\frac{1}{3}}=x^0=1\).

\(1\)