Free math worksheets

- For a monic trinomial, focus on the constant term and the coefficient of \(x\). - Look for two integers that meet both the product and sum conditions. - Multiply your factors to check the middle term.

1. Look for two integers whose product is \(10\) and whose sum is \(7\). 2. The numbers are \(5\) and \(2\), so \(x^2+7x+10=(x+5)(x+2)\).

\((x+5)(x+2)\)

- First factor out the greatest common factor. - Then factor the remaining quadratic trinomial. - Check your result by multiplying the factors.

1. Factor out the greatest common factor: \(45n^2 - 60n + 15 = 15(3n^2 - 4n + 1)\). 2. Factor the trinomial: \(3n^2 - 4n + 1 = (3n - 1)(n - 1)\). 3. Therefore, \(45n^2 - 60n + 15 = 15(3n - 1)(n - 1)\).

\(15(3n - 1)(n - 1)\)

- Look for a common numerical factor first. - Consider factoring out a negative common factor so the leading term inside the parentheses is positive. - Check your result by multiplying the factors.

1. Factor out \(-4\): \(-8b^2 + 20b + 12 = -4(2b^2 - 5b - 3)\). 2. Factor the trinomial: \(2b^2 - 5b - 3 = (2b + 1)(b - 3)\). 3. Therefore, \(-8b^2 + 20b + 12 = -4(2b + 1)(b - 3)\).

\(-4(2b + 1)(b - 3)\)

- First factor out the greatest common factor. - Then factor the remaining quadratic trinomial. - Multiply the factors to check your work.

1. Factor out the greatest common factor: \(12z^2 - 30z + 12 = 6(2z^2 - 5z + 2)\). 2. Factor the trinomial: \(2z^2 - 5z + 2 = (2z - 1)(z - 2)\). 3. Therefore, \(12z^2 - 30z + 12 = 6(2z - 1)(z - 2)\).

\(6(2z - 1)(z - 2)\)

- Begin by factoring out the greatest common factor. - Use the signs of the constant and middle terms to choose the signs in the binomial factors. - Check by multiplying the factors.

1. Factor out the greatest common factor: \(20y^2 + 10y - 30 = 10(2y^2 + y - 3)\). 2. Factor the trinomial: \(2y^2 + y - 3 = (2y + 3)(y - 1)\). 3. Therefore, \(20y^2 + 10y - 30 = 10(2y + 3)(y - 1)\).

\(10(2y + 3)(y - 1)\)

- Factor out the greatest common factor first. - Then find two binomial factors whose product gives the remaining trinomial. - Make sure the expression is fully factored, and check by multiplying.

1. Factor out the greatest common factor: \(18x^2 - 12x - 6 = 6(3x^2 - 2x - 1)\). 2. Factor the trinomial: \(3x^2 - 2x - 1 = (3x + 1)(x - 1)\). 3. Therefore, \(18x^2 - 12x - 6 = 6(3x + 1)(x - 1)\).

\(6(3x + 1)(x - 1)\)

- Move all terms to one side. - Recognize a difference of squares. - Equal function outputs correspond to graph intersections.

a) Rewrite with all terms on one side: \(x^2 - 4 = 0\). Factor the difference of squares: \((x - 2)(x + 2) = 0\). b) By the zero-product property, \(x = -2\) or \(x = 2\), so the solution set is \(\{-2, 2\}\). c) These values are the x-coordinates of the intersection points of the graphs of \(f\) and \(g\).

a) \(x^2 - 4 = 0\), so \((x - 2)(x + 2) = 0\) b) \(\{-2, 2\}\) c) They are the x-coordinates of the graphs’ intersection points.

- Equal outputs mean the function expressions are equal. - Move all terms to one side. - Factor the resulting quadratic.

1. Set the function expressions equal: \(x^2 = 2x + 8\). 2. Rearrange: \(x^2 - 2x - 8 = 0\). 3. Factor: \((x - 4)(x + 2) = 0\). 4. Therefore, \(x = -2\) or \(x = 4\).

\(x = -2\) and \(x = 4\)

- Move all terms to one side and factor. - An equation \(f(x) = g(x)\) identifies equal outputs. - Equal outputs correspond to graph intersections.

1. Move all terms to one side: \(x^2 - x - 6 = 0\). 2. Factor: \((x - 3)(x + 2) = 0\). 3. Thus, \(x = -2\) or \(x = 3\). 4. These values are the x-coordinates of the intersection points of the graphs of \(f\) and \(g\).

a) \(\{-2, 3\}\) b) The solutions are the x-coordinates of the graphs’ intersection points.

- Look for the greatest common factor that includes the largest possible power of \(x\). - Check your factorization by multiplying before you use the zero-product property. - A product can equal zero only when at least one factor equals zero.

1. For a), factor out \(7x\): \(7x(2x - 3)=0\). By the zero-product property, \(x=0\) or \(x=\frac{3}{2}\). 2. For b), factor out \(0.5x\): \(0.5x(x+8)=0\). By the zero-product property, \(x=0\) or \(x=-8\).

a) \(7x(2x-3)=0\); \(x\in\left\{0,\frac{3}{2}\right\}\) b) \(0.5x(x+8)=0\); \(x\in\{-8,0\}\)

- Equal outputs mean the function expressions are equal. - Clear the decimal coefficient before factoring. - Substitute each x-value into either function to find the corresponding y-value.

1. Set the function expressions equal: \(x^2 = 1.5x + 1\). 2. Multiply by \(2\): \(2x^2 - 3x - 2 = 0\). 3. Factor: \((2x + 1)(x - 2) = 0\). 4. Thus, \(x = -0.5\) or \(x = 2\). 5. The corresponding outputs are \(f(-0.5) = 0.25\) and \(f(2) = 4\), so the intersection points are \((-0.5, 0.25)\) and \((2, 4)\).

\(x = -0.5\) or \(x = 2\); intersections: \((-0.5, 0.25)\) and \((2, 4)\)

- Equal outputs mean the function expressions are equal. - Clear the decimal coefficient before factoring. - Substitute each candidate value into both sides of the original equation.

1. Set the function expressions equal: \(0.5x^2 = x + 4\). 2. Multiply by \(2\): \(x^2 - 2x - 8 = 0\). 3. Factor: \((x - 4)(x + 2) = 0\), so \(x = -2\) or \(x = 4\). 4. Check \(x = -2\): \(0.5 \cdot (-2)^2 = 2\) and \(-2 + 4 = 2\). 5. Check \(x = 4\): \(0.5 \cdot 4^2 = 8\) and \(4 + 4 = 8\). 6. Both values satisfy the equation.

\(\{-2, 4\}\)

- Set the function expressions equal. - Isolate \(x^2\). - Remember both square roots of a positive number.

1. Set the functions equal: \(x^2 - 2 = -x^2 + 6\). 2. Simplify: \(2x^2 = 8\), so \(x^2 = 4\). 3. Therefore, \(x = -2\) or \(x = 2\). 4. Substituting either value into \(f\) gives \(y = 2\). 5. The intersection points are \((-2, 2)\) and \((2, 2)\).

\((-2, 2)\) and \((2, 2)\)

- Rewrite the constant as a perfect square. - Look for a product of two conjugate binomials. - Use the zero-product property only after the equation is factored.

1. Recognize a difference of squares: \(x^2-16=x^2-4^2\). 2. Factor: \((x-4)(x+4)=0\). 3. Apply the zero-product property to obtain \(x=4\) or \(x=-4\).

\((x-4)(x+4)=0\); \(x\in\{-4,4\}\)

- For each expression, decide first whether a common factor can be removed. - When a trinomial remains, look for numbers whose product and sum match its coefficients. - For the expression containing only \(x^2\) and a positive constant, think about the possible values of a real square.

1. For a), factor out \(2\), then factor the trinomial: \(p(x)=2(x^2-6x+5)=2(x-1)(x-5)\). Thus the real zeros are \(x=1\) and \(x=5\). 2. For b), \(x^2+4>0\) for every real \(x\), so it has no real zeros and cannot be written as a product of real linear factors. 3. For c), factor out the greatest common factor: \(r(x)=5x(x-2)\). Thus the real zeros are \(x=0\) and \(x=2\).

a) \(p(x)=2(x-1)(x-5)\); zeros \(x=1,5\) b) No product of real linear factors; no real zeros because \(x^2+4>0\) for every real \(x\). c) \(r(x)=5x(x-2)\); zeros \(x=0,2\)

- First decide what factor structure is already visible in each function. - A missing constant term often signals a common factor of \(x\). - A subtraction of two perfect squares can be factored into conjugates.

1. a) Set the factored expression equal to zero: \((x-6)(x+4)=0\). The zeros are \(x=6\) and \(x=-4\). 2. b) Factor out \(x\): \(x(x-5)=0\). The zeros are \(x=0\) and \(x=5\). 3. c) Factor the trinomial: \((x-6)(x+2)=0\). The zeros are \(x=6\) and \(x=-2\). 4. d) Factor the difference of squares: \(4(x-2)(x+2)=0\). The zeros are \(x=2\) and \(x=-2\).

a) \((x-6)(x+4)=0\); zeros \(x=-4,6\) b) \(x(x-5)=0\); zeros \(x=0,5\) c) \((x-6)(x+2)=0\); zeros \(x=-2,6\) d) \(4(x-2)(x+2)=0\); zeros \(x=-2,2\)

- Collect all terms on one side before looking for a common factor. - In each result, every term contains \(x\). - Verify the factorization before applying the zero-product property.

1. a) Rearrange to \(4x^2-12x=0\). Factor: \(4x(x-3)=0\), so \(x=0\) or \(x=3\). 2. b) Rearrange to \(-x^2+5x=0\). Factor: \(-x(x-5)=0\), so \(x=0\) or \(x=5\).

a) \(4x(x-3)=0\); \(x\in\{0,3\}\) b) \(-x(x-5)=0\); \(x\in\{0,5\}\)

- Treat the two sides as separate functions. - Equal outputs give an equation that can be factored. - Solutions of \(f(x) = g(x)\) correspond to graph intersections.

1. Set the function expressions equal: \(x^2 = -2x + 3\). 2. Move all terms to one side: \(x^2 + 2x - 3 = 0\). 3. Factor: \((x + 3)(x - 1) = 0\). 4. Therefore, \(x = -3\) or \(x = 1\). 5. These values are the x-coordinates of the points where the graphs of \(f\) and \(g\) intersect.

1. \(x = -3\) or \(x = 1\) 2. They are the x-coordinates of the graphs’ intersection points.

- Find a factor pair of \(-21\) whose sum is \(-4\). - Set each linear factor equal to zero. - Compare a zero \(r\) with its factor \(x - r\).

1. Find two numbers whose product is \(-21\) and whose sum is \(-4\): \(-7\) and \(3\). Therefore, \(x^2 - 4x - 21 = (x - 7)(x + 3)\). 2. The zeros are \(7\) and \(-3\). A zero \(r\) corresponds to the factor \(x-r\), so the zero \(-3\) gives \(x-(-3)=x+3\). Thus, the sign of the number written in a linear factor is opposite the sign of its zero.

1. \((x - 7)(x + 3)\) 2. Each zero \(r\) gives a factor \(x - r\), so the sign of the number in the factor is opposite the sign of the zero.

- Put the second equation in zero-product form before factoring. - Identify the greatest common numerical factor and the common power of \(x\). - Multiply the factors back together to check them.

1. a) Factor: \(7x(x+3)=0\), so \(x=0\) or \(x=-3\). 2. b) Rearrange to \(0.5x^2-2x=0\), then factor: \(0.5x(x-4)=0\), so \(x=0\) or \(x=4\).

a) \(7x(x+3)=0\); \(x\in\{-3,0\}\) b) \(0.5x(x-4)=0\); \(x\in\{0,4\}\)

- There is no common factor to remove first. - For a nonmonic trinomial, use the product of the leading coefficient and constant to plan a middle-term split. - After splitting the middle term, look for a common binomial factor.

1. The product of the leading coefficient and constant is \(6\cdot(-2)=-12\). Two integers with product \(-12\) and sum \(1\) are \(4\) and \(-3\). 2. Split the middle term: \(6x^2+4x-3x-2\). 3. Factor by grouping: \(2x(3x+2)-(3x+2)=(3x+2)(2x-1)\).

\((3x+2)(2x-1)\)

- Check whether the first and last terms are perfect squares. - For three terms, compare the middle term with twice the product of the square roots of the outer terms. - For two squared terms separated by subtraction, think about conjugate factors.

1. In a), \(36=6^2\) and the middle term is \(-2\cdot6\cdot x\), so the expression is the perfect-square trinomial \((x-6)^2\). 2. In b), \(49y^2=(7y)^2\) and \(16=4^2\), so the difference of squares factors as \((7y-4)(7y+4)\).

a) \((x-6)^2\); perfect-square trinomial b) \((7y-4)(7y+4)\); difference of squares

- Do not assume every quadratic with integer coefficients factors over the integers. - For the monic expression, test integer factor pairs of the constant against the middle coefficient. - For the nonmonic expression, look for a middle-term split that supports grouping.

1. For a), the integer factor pairs of \(7\) have sums \(8\) or \(-8\), not \(5\). Therefore, it does not factor into integer-coefficient binomials. 2. For b), split the middle term: \(2x^2+6x+x+3\). 3. Factor by grouping: \(2x(x+3)+1(x+3)=(2x+1)(x+3)\).

a) Not factorable over the integers. b) \((2x+1)(x+3)\)

- Test the value that makes \(x - 4\) equal to zero in the original equation. - Ask when division by \(x - 4\) is not allowed. - Move all terms to one side and look for a common binomial factor.

a) Solving \(x - 4 = 3\) gives \(x = 7\). b) Dividing by \(x - 4\) assumes \(x - 4 \ne 0\). This excludes \(x = 4\), even though \(x = 4\) makes both sides of the original equation equal to zero. c) Move all terms to one side: \((x - 4)^2 - 3(x - 4) = 0\). Factor: \((x - 4)((x - 4) - 3) = (x - 4)(x - 7) = 0\). Thus, \(x = 4\) or \(x = 7\).

a) \(x = 7\) b) Dividing by \(x - 4\) discards the case \(x - 4 = 0\). Division by zero is undefined, and \(x = 4\) is a solution of the original equation. c) \(x \in \{4, 7\}\)

- Decide whether a greatest common factor, a difference of squares, or trinomial factoring fits each equation. - Clearing the decimal in part c) can make the integer factor pair easier to see. - Do not state roots until you have written the product that equals zero.

1. a) Factor out \(2x\): \(2x(x-9)=0\), so \(x=0\) or \(x=9\). 2. b) Factor out \(3\) and use a difference of squares: \(3((x-1)^2-9)=3(x-4)(x+2)=0\), so \(x=-2\) or \(x=4\). 3. c) Multiply the equation by \(2\): \(x^2-2x-8=0\). Factor: \((x-4)(x+2)=0\), so \(x=-2\) or \(x=4\). 4. d) The equation is already factored. The zero-product property gives \(2x-4=0\) or \(x+5=0\), so \(x=2\) or \(x=-5\).

a) \(2x(x-9)=0\); \(x\in\{0,9\}\) b) \(3(x-4)(x+2)=0\); \(x\in\{-2,4\}\) c) \((x-4)(x+2)=0\); \(x\in\{-2,4\}\) d) \((2x-4)(x+5)=0\); \(x\in\{-5,2\}\)

- Check whether the constant is a perfect square and the middle term is missing. - Check whether every term has a common factor. - For a trinomial, look for two numbers whose product is the constant term and whose sum is the coefficient of \(x\).

1. The first equation is a difference of squares: \(x^2 - 144 = (x - 12)(x + 12)\). Therefore, \(x = -12\) or \(x = 12\). 2. The second equation has a greatest common factor of \(2x\): \(2x(x + 5) = 0\). Therefore, \(x = -5\) or \(x = 0\). 3. The third equation is a factorable trinomial: \(x^2 - 2x - 15 = (x - 5)(x + 3)\). Therefore, \(x = -3\) or \(x = 5\).

1. Difference of squares; \(x \in \{-12, 12\}\) 2. Greatest common factor; \(x \in \{-5, 0\}\) 3. Trinomial factoring; \(x \in \{-3, 5\}\)

- Set the function expressions equal to find intersections. - Factor the resulting quadratic. - Test a point by substituting its x-coordinate into each function.

1. Set the functions equal: \(x^2 - 4x + 5 = x + 1\). 2. Rearrange and factor: \(x^2 - 5x + 4 = (x - 1)(x - 4) = 0\). 3. Thus, \(x = 1\) or \(x = 4\). Using \(g(x) = x + 1\), the corresponding y-values are \(2\) and \(5\). The intersections are \((1, 2)\) and \((4, 5)\). 4. For \(Q\), \(g(2.5) = 3.5\), so \(Q\) lies on the line. 5. However, \(p(2.5) = 2.5^2 - 4 \cdot 2.5 + 5 = 1.25\), not \(3.5\). Therefore, \(Q\) does not lie on the parabola.

a) \((1, 2)\) and \((4, 5)\) b) \(Q\) lies on the line only.

- Express the longer leg using the variable for the shorter leg. - Substitute both leg expressions into the Pythagorean theorem before simplifying. - After factoring, check which algebraic solution can represent a physical length.

1. The longer leg is \(a+3\), so the Pythagorean theorem gives \(a^2+(a+3)^2=29\). 2. Expand and simplify: \(2a^2+6a+9=29\), so \(2a^2+6a-20=0\). 3. Divide by \(2\): \(a^2+3a-10=0\). 4. Factor: \((a+5)(a-2)=0\). 5. The algebraic solutions are \(a=-5\) and \(a=2\). A length must be positive, so the shorter leg is \(2\,\text{cm}\) and the longer leg is \(5\,\text{cm}\).

Standard form: \(a^2+3a-10=0\) Factorization: \((a+5)(a-2)=0\) Leg lengths: \(2\,\text{cm}\) and \(5\,\text{cm}\)

- Move all terms to one side and simplify before deciding whether factoring applies. - Check whether the simplified expression is a difference of squares. - If a quadratic is a sum of positive squares over the reals, consider whether real linear factors can exist.

1. a) Simplify to \(x^2-9=0\). Factor: \((x-3)(x+3)=0\), so \(x=-3\) or \(x=3\). 2. b) Simplify to \(x^2+27=0\). It has no factorization into real linear factors and no real solution. 3. c) Expand and rearrange to \(x^2-36=0\). Factor: \((x-6)(x+6)=0\), so \(x=-6\) or \(x=6\). 4. d) Divide by \(0.5\) to obtain \(x^2-0.01=0\). Factor: \((x-0.1)(x+0.1)=0\), so \(x=-0.1\) or \(x=0.1\).

a) \(x^2-9=0=(x-3)(x+3)\); \(x\in\{-3,3\}\) b) \(x^2+27=0\); no real linear factorization; \(\varnothing\) c) \(x^2-36=0=(x-6)(x+6)\); \(x\in\{-6,6\}\) d) \(x^2-0.01=0=(x-0.1)(x+0.1)\); \(x\in\{-0.1,0.1\}\)

- Clear the denominator by multiplying by \(p\). - Recognize the resulting difference of squares. - Use the condition \(p \ne 0\) to justify that the two roots are different.

a) When \(p=5\), the equation is \(\frac{x^2}{5}-5=0\). Multiplying by \(5\) gives \(x^2-25=0\), so \((x-5)(x+5)=0\). Thus, \(x=-5\) or \(x=5\). b) For \(p\ne0\), multiply by \(p\): \(x^2-p^2=0\). Then \((x-p)(x+p)=0\), so \(x=p\) or \(x=-p\). Because \(p\ne0\), these two values are distinct. c) The solution set is \(\{-p,p\}\).

a) \(x \in \{-5, 5\}\) b) \((x - p)(x + p) = 0\), so the two distinct solutions are \(x = p\) and \(x = -p\). c) \(x \in \{-p, p\}\)

- Combine like terms before deciding what to factor. - Clearing fractions is optional if you can see the common factor directly. - A zero constant term is a cue to check for a factor of \(x\).

1. a) Move terms left: \(\frac{1}{2}x^2+3x=0\). Factor: \(x\left(\frac{1}{2}x+3\right)=0\), so \(x=0\) or \(x=-6\). 2. b) Expand and simplify to \(x^2+5x=0\). Factor: \(x(x+5)=0\), so \(x=0\) or \(x=-5\).

a) \(x\left(\frac{1}{2}x+3\right)=0\); \(x\in\{-6,0\}\) b) \(x(x+5)=0\); \(x\in\{-5,0\}\)

- Simplify both sides far enough to expose a recognizable quadratic pattern. - Watch for cancellation of the linear terms. - Once you have a difference of squares, write the two conjugate factors explicitly.

1. Expand to \(x^2-12x=64-12x\), so \(x^2-64=0\). Factor: \((x-8)(x+8)=0\), giving \(x=-8\) or \(x=8\). 2. Use differences of squares: \(x^2-9+x^2-16=7\). Simplify to \(x^2-16=0\). Factor: \((x-4)(x+4)=0\), giving \(x=-4\) or \(x=4\).

1) \((x-8)(x+8)=0\); \(x\in\{-8,8\}\) 2) \((x-4)(x+4)=0\); \(x\in\{-4,4\}\)

- Expand only as much as needed to collect every term on one side. - For the first equation, look for a common factor after simplification. - For the second, find two integers with the required product and sum.

1. Expand and simplify: \(2x^2-10x=0\). Factor: \(2x(x-5)=0\), so \(x=0\) or \(x=5\). 2. Expand and rearrange: \(x^2-7x+6=0\). Factor: \((x-1)(x-6)=0\), so \(x=1\) or \(x=6\).

1) \(2x(x-5)=0\); \(x\in\{0,5\}\) 2) \((x-1)(x-6)=0\); \(x\in\{1,6\}\)

- Distribute the subtraction across the entire squared binomial before combining terms. - Expect substantial cancellation after both sides are expanded. - Once only \(x^2\) and a constant remain, check for a difference of squares.

1. Expand: \(9x^2-6x+1-(x^2+4x+4)+4x^2-25=4x^2-10x+4\). 2. Combine like terms: \(12x^2-10x-28=4x^2-10x+4\). 3. Rearrange: \(8x^2-32=0\), then divide by \(8\) to get \(x^2-4=0\). 4. Factor: \((x-2)(x+2)=0\). 5. Therefore, \(x=-2\) or \(x=2\).

Simplified equation: \(x^2-4=0\) Factorization: \((x-2)(x+2)=0\) Solution set: \(x\in\{-2,2\}\)

- Expand each square and product carefully before comparing the two sides. - After collecting terms on one side, identify the integer pair that matches the trinomial. - Check the binomial product before using the zero-product property.

1. Expand both sides: \(x^2-6x+9+x^2+4x+4=x^2-16+2x+34\). 2. Combine like terms: \(2x^2-2x+13=x^2+2x+18\). 3. Rearrange: \(x^2-4x-5=0\). 4. Factor: \((x-5)(x+1)=0\). 5. Therefore, \(x=5\) or \(x=-1\).

\(x^2-4x-5=0=(x-5)(x+1)\); \(x\in\{-1,5\}\)

- Both equations already give \(y\) in terms of \(x\). - Set the right sides equal and factor. - Substitute each x-value to find its paired y-value.

1. Both equations are solved for \(y\), so their right sides must be equal at an intersection. 2. Set them equal: \(x^2 - 4x + 1 = 2x - 4\). 3. Rearrange and factor: \(x^2 - 6x + 5 = (x - 1)(x - 5) = 0\). 4. Thus, \(x = 1\) or \(x = 5\). 5. Substitute into \(y = 2x - 4\): the corresponding y-values are \(-2\) and \(6\). 6. The solutions are \((1, -2)\) and \((5, 6)\).

\(\{(1, -2), (5, 6)\}\)

- Solve the linear equation for one variable. - Substitute that expression into the squared equation. - Remember both cases when taking a square root.

1. Solve the linear equation for \(x\): \(x = y + 1\). 2. Substitute into the first equation: \((y + 1 + y)^2 = 25\), so \((2y + 1)^2 = 25\). 3. Take both square roots: \(2y + 1 = 5\) or \(2y + 1 = -5\). 4. The first case gives \(y = 2\) and \(x = 3\). The second case gives \(y = -3\) and \(x = -2\). 5. Therefore, the system has exactly two ordered-pair solutions.

\(\{(3, 2), (-2, -3)\}\)

- Solve the linear equation for one variable. - Substitute into the quadratic equation and keep the parentheses. - Factor the resulting quadratic, then find each matching value of the other variable.

1. Solve the linear equation for \(y\): \(y = 3 - x\). 2. Substitute into the quadratic equation: \(3x^2 - (3 - x)^2 = 11\). 3. Expand and simplify: \(3x^2 - (9 - 6x + x^2) = 11\), so \(2x^2 + 6x - 20 = 0\). 4. Divide by \(2\): \(x^2 + 3x - 10 = 0\). 5. Factor: \((x - 2)(x + 5) = 0\). Thus, \(x = 2\) or \(x = -5\). 6. Using \(y = 3 - x\), the corresponding values are \(y = 1\) and \(y = 8\).

\(\{(2, 1), (-5, 8)\}\)

- Solve the linear equation for one variable. - Substitute into the quadratic equation and expand carefully. - Factor the resulting quadratic, then find the matching value of the other variable.

1. Solve the linear equation for \(x\): \(x = 3 - y\). 2. Substitute into the quadratic equation: \((3 - y)^2 + 2y^2 = 33\). 3. Expand and simplify: \(9 - 6y + 3y^2 = 33\), so \(y^2 - 2y - 8 = 0\). 4. Factor: \((y - 4)(y + 2) = 0\). Thus, \(y = 4\) or \(y = -2\). 5. Using \(x = 3 - y\), the corresponding values are \(x = -1\) and \(x = 5\).

\(\{(-1, 4), (5, -2)\}\)

- Solve the sum equation for one variable. - Substitute into the product equation. - Factor the resulting quadratic and find each corresponding ordered pair.

1. Solve the first equation for \(y\): \(y = -2 - x\). 2. Substitute into the product equation: \(x(-2 - x) = -15\). 3. Rewrite in standard form: \(x^2 + 2x - 15 = 0\). 4. Factor: \((x + 5)(x - 3) = 0\). Thus, \(x = -5\) or \(x = 3\). 5. The corresponding values of \(y\) are \(3\) and \(-5\).

\(\{(-5, 3), (3, -5)\}\)

- Set the two expressions for \(y\) equal. - Factor the resulting quadratic. - Substitute each x-value into either equation to find the paired y-value.

1. Set the function expressions equal: \(x^2 - 3 = 2x\). 2. Rearrange and factor: \(x^2 - 2x - 3 = (x - 3)(x + 1) = 0\). 3. Thus, \(x = -1\) or \(x = 3\). 4. Substituting into \(y = 2x\) gives \(y = -2\) and \(y = 6\). 5. The solutions are \((-1, -2)\) and \((3, 6)\); these are the graphs’ intersection points.

\((-1, -2)\) and \((3, 6)\)

- Factor out \(x\). - Check when the two solution expressions coincide. - For part b, substitute the given value of \(x\) into the original equation.

a) Factor: \(x(3x+p)=0\). Therefore, \(x=0\) or \(x=-\frac{p}{3}\). When \(p=0\), these expressions give the same solution, \(x=0\); when \(p\ne0\), they are distinct. b) Substitute \(x=-4\) into the original equation: \(3(-4)^2+p(-4)=0\). Then \(48-4p=0\), so \(p=12\).

a) For \(p \ne 0\), \(x \in \left\{0, -\frac{p}{3}\right\}\). For \(p = 0\), \(x = 0\). b) \(p = 12\)

- Read the intersection coordinates from the axes. - Use point-slope information for the line and vertex form for the parabola. - Set the equations equal and factor.

1. From the graph, the intersection points are \((-1, 0)\) and \((4, 5)\). 2. Write \(g(x) = x + b\). Substituting \((-1, 0)\) gives \(b = 1\), so \(g(x) = x + 1\). 3. Write \(f(x) = a(x - 1)^2 - 4\). Substituting \((3, 0)\) gives \(0 = 4a - 4\), so \(a = 1\). 4. Therefore, \(f(x) = (x - 1)^2 - 4 = x^2 - 2x - 3\). 5. Set the functions equal: \(x^2 - 2x - 3 = x + 1\), so \(x^2 - 3x - 4 = 0\). 6. Factor: \((x - 4)(x + 1) = 0\), giving \(x = -1\) or \(x = 4\). 7. Substitution into \(g\) gives the points \((-1, 0)\) and \((4, 5)\), matching the graph.

1. \((-1, 0)\) and \((4, 5)\) 2. \(g(x) = x + 1\), \(f(x) = x^2 - 2x - 3\) 3. The calculated intersections are \((-1, 0)\) and \((4, 5)\).

- Identify which two graphs meet at each labeled point. - Set the corresponding function expressions equal. - Factor the quadratic equations.

1. From the graph, \(g(x) = -x + 2\) and \(h(x) = x\). 2. Points \(A\) and \(D\) are intersections of \(p\) and \(g\), so solve \(x^2 - 2x - 4 = -x + 2\). 3. This becomes \(x^2 - x - 6 = (x - 3)(x + 2) = 0\), so the x-coordinates are \(-2\) and \(3\). 4. Points \(B\) and \(E\) are intersections of \(p\) and \(h\), so solve \(x^2 - 2x - 4 = x\). 5. This becomes \(x^2 - 3x - 4 = (x - 4)(x + 1) = 0\), so the x-coordinates are \(-1\) and \(4\). 6. Point \(C\) is the intersection of \(g\) and \(h\), so solve \(-x + 2 = x\), giving \(x = 1\).

a) \(A,D\): \(x^2 - 2x - 4 = -x + 2\); \(B,E\): \(x^2 - 2x - 4 = x\); \(C\): \(-x + 2 = x\) b) \(A,D\): \(x = -2, 3\); \(B,E\): \(x = -1, 4\); \(C\): \(x = 1\)

- Read both coordinates at each crossing. - Set the function expressions equal. - Factor out \(x\).

1. From the graph, the intersections are \((0, 0)\) and \((3, 3)\). 2. Set the functions equal: \(-x^2 + 4x = x\). 3. Rearrange and factor: \(x^2 - 3x = x(x - 3) = 0\). 4. Thus, \(x = 0\) or \(x = 3\). 5. Since \(g(x) = x\), the corresponding y-values are \(0\) and \(3\), confirming the graph.

a) \((0, 0)\) and \((3, 3)\) b) The calculation gives the same points.

- Look for the x-values where the parabola and line meet. - Consider what happens when two equal values are subtracted. - Recall the name for an x-value where a function's output is \(0\).

1. From the graph, the functions intersect at \(x_1=2\) and \(x_2=4\). 2. Compute the difference: \(h(x)=f(x)-g(x)=0.5x^2-3x+4\). 3. At each intersection, \(f(x)=g(x)\). Therefore, \(h(2)=f(2)-g(2)=0\) and \(h(4)=f(4)-g(4)=0\). Thus, \(2\) and \(4\) are the zeros of \(h\). 4. In general, \(f(x)=g(x)\) is equivalent to \(f(x)-g(x)=0\), so the x-coordinates of the intersections of two graphs are the zeros of their difference function.

a) \(x_1=2\) and \(x_2=4\) b) The values \(2\) and \(4\) are the zeros of \(h\). c) At an intersection, \(f(x)=g(x)\), which is equivalent to \(f(x)-g(x)=0\).

- Use the vertex and another visible point to determine the parabola. - Use two points to find the line's slope and y-intercept. - At an intersection, the function outputs are equal. - Rearrange the quadratic equation and look for a factorization.

1. The parabola has vertex \((0,4)\), opens downward, and passes through \((2,0)\). Therefore, \(f(x)=-x^2+4\). 2. The line has y-intercept \(2\) and slope \(-1\), so \(g(x)=-x+2\). 3. Set the functions equal: \(-x^2+4=-x+2\). 4. Rearrange to obtain \(x^2-x-2=0\), and factor: \((x-2)(x+1)=0\). Thus, \(x=2\) or \(x=-1\). 5. Substitute into \(g\). The corresponding points are \((2,0)\) and \((-1,3)\), which agree with the graph.

a) \(f(x)=-x^2+4\); \(g(x)=-x+2\) b) \(-x^2+4=-x+2\) c) \((-1,3)\) and \((2,0)\)

- At an intersection, the two functions have equal outputs. - Move all terms to one side and factor the quadratic expression. - An intersection point includes both an x-coordinate and a y-coordinate.

1. At an intersection, \(f(x)=g(x)\), so \(x^2-4=x-2\). 2. Move all terms to one side: \(x^2-x-2=0\). 3. Factor: \((x-2)(x+1)=0\). Therefore, \(x=2\) or \(x=-1\). 4. Substitute into \(g(x)=x-2\). When \(x=2\), \(y=0\). When \(x=-1\), \(y=-3\). 5. The intersections are \((2,0)\) and \((-1,-3)\), which match the graph.

a) \(x^2-4=x-2\) b) \((-1,-3)\) and \((2,0)\) c) The calculated points agree with the graph.

- Connect the zeros of a function to an equation of the form \(f(x)=0\). - In the second graph, interpret what it means when the two function outputs are equal. - Decide whether x-coordinates or y-coordinates represent the equation solutions. - Find two integers whose product is \(-2\) and whose sum is \(1\).

1. In graph a), the function \(f(x)=x^2+x-2\) is shown. The solutions are the x-intercepts of the parabola because they satisfy \(f(x)=0\). 2. In graph b), the equation is rewritten as \(x^2=-x+2\). The solutions are the x-coordinates of the intersections of \(y=x^2\) and \(y=-x+2\). 3. Both graphs show the solutions \(x=-2\) and \(x=1\). 4. To verify, factor the original equation: \(x^2+x-2=(x+2)(x-1)=0\). Therefore, \(x=-2\) or \(x=1\), confirming the graphical results.

a) In graph a), use the x-intercepts of \(f\). In graph b), use the x-coordinates of the intersections of the parabola and line. b) \(x=-2\) and \(x=1\) c) \((x+2)(x-1)=0\), so \(x=-2\) or \(x=1\).

- At an intersection, the two functions have equal outputs. - Move all terms to one side so the other side is \(0\). - Look for two factors of the quadratic expression. - Find the y-coordinate for each solution.

1. Set the functions equal: \(x^2-2x-1=x-3\). 2. Rearrange: \(x^2-3x+2=0\). 3. Factor: \((x-1)(x-2)=0\), so \(x=1\) or \(x=2\). 4. Substitute into \(g\): \(g(1)=-2\) and \(g(2)=-1\). 5. Therefore, the intersections are \((1,-2)\) and \((2,-1)\), which agree with the graph.

\((1,-2)\) and \((2,-1)\)

- Check a proposed factorization by multiplying it back out. - Compare all three coefficients of the product with the original trinomial. - Refactor the original expression rather than changing only one guessed factor.

1. Expanding Maya's factors gives \(6x^2+10x+4\), which does not match the original trinomial. 2. Split the middle term in the original expression: \(6x^2+4x+3x+2\). 3. Factor by grouping: \(2x(3x+2)+1(3x+2)=(3x+2)(2x+1)\).

Maya's factors expand to \(6x^2+10x+4\), not \(6x^2+7x+2\). The correct factorization is \((3x+2)(2x+1)\).

- Solve the sum equation for one variable. - Substitute into the product equation and factor the resulting quadratic. - Determine when the two ordered-pair expressions are identical.

1. Solve the first equation for \(y\): \(y = 2a + b - x\). 2. Substitute into the product equation: \(x(2a + b - x) = a^2 + ab\). 3. Rewrite in standard form: \(x^2 - (2a + b)x + a^2 + ab = 0\). 4. Factor: \((x - a)(x - a - b) = 0\). Thus, \(x = a\) or \(x = a + b\). 5. The corresponding values of \(y\) are \(a + b\) and \(a\). When \(b = 0\), the two ordered-pair expressions coincide.

For \(b \ne 0\), \(\{(a, a + b), (a + b, a)\}\). For \(b = 0\), \(\{(a, a)\}\).

- Use the larger square's area to determine \(a+b\). - The inner-square side is the hypotenuse of each corner triangle, giving a second relation between \(a\) and \(b\). - Substitute one leg as the larger-square side minus the other, then factor the resulting quadratic.

1. The larger square has side length \(\sqrt{81}=9\,\text{cm}\), so \(a+b=9\). 2. Each corner triangle has hypotenuse \(\sqrt{53}\,\text{cm}\), so \(a^2+b^2=53\). 3. Set \(b=9-a\): \(a^2+(9-a)^2=53\). 4. Simplify: \(2a^2-18a+28=0\), so \(a^2-9a+14=0\). 5. Factor: \((a-2)(a-7)=0\). Thus the two leg lengths are \(2\,\text{cm}\) and \(7\,\text{cm}\), in either order.

Quadratic: \(x^2-9x+14=0\) Factorization: \((x-2)(x-7)=0\) \(\{a,b\}=\{2\,\text{cm},7\,\text{cm}\}\)