Free math worksheets

- Identify the value \(c\) from the divisor \(x-c\). - The remainder theorem replaces the division calculation with one function evaluation. - Substitute carefully into every term.

1. For divisor \(x-2\), the remainder theorem says the remainder is \(f(2)\). 2. Compute \(f(2)=2(2)^3-(2)^2+4(2)-5=15\).

The remainder is \(15\).

- A divisor \(x-c\) is a factor exactly when the polynomial evaluates to \(0\) at \(c\). - Pay attention to the sign in \(x+3\) when identifying the test value. - You do not need to calculate a quotient.

a) For \(x+3=x-(-3)\), evaluate \(p(-3)=-27+18+15-6=0\). By the factor theorem, \(x+3\) is a factor. b) For \(x-1\), evaluate \(p(1)=1+2-5-6=-8\ne0\). Therefore, \(x-1\) is not a factor.

a) Yes, \(x+3\) is a factor. b) No, \(x-1\) is not a factor.

- Rewrite the divisor in the form \(x-c\) before choosing the input value. - The sign of \(c\) matters when the divisor is \(x+2\). - Evaluate the polynomial directly instead of dividing it.

1. Since \(x+2=x-(-2)\), the remainder is \(g(-2)\). 2. Compute \(g(-2)=(-2)^4-3(-2)^3+2(-2)-7=16+24-4-7=29\).

The remainder is \(29\).

- Rewrite each divisor in the form \(x-c\). - The sign of \(c\) is opposite the visible constant in a divisor such as \(x+2\). - A zero remainder also tells you something about factor status.

a) For \(x-2\), evaluate \(f(2)=8-8-10+6=-4\). The remainder is \(-4\). b) Since \(x+2=x-(-2)\), evaluate \(f(-2)=-8-8+10+6=0\). The remainder is \(0\), so \(x+2\) is a factor. c) The remainder theorem uses the value \(c\) in a divisor written as \(x-c\). Thus, \(x-2\) uses \(c=2\), while \(x+2\) uses \(c=-2\).

a) \(-4\) b) \(0\) c) \(x-2=x-(2)\), but \(x+2=x-(-2)\), so the evaluation inputs are \(2\) and \(-2\).

- Translate “\(x-2\) is a factor” into a statement about \(p(2)\). - Substitute before solving for the unknown coefficient. - The factor theorem turns the factor condition into one linear equation in \(k\).

1. If \(x-2\) is a factor, then \(p(2)=0\). 2. Compute \(p(2)=8+4k-10+6=4k+4\). 3. Solve \(4k+4=0\) to get \(k=-1\).

\(k=-1\)

- For division by \(x-c\), the remainder theorem connects the remainder directly to \(P(c)\). - A linear expression \(x-c\) is a factor exactly when the corresponding function value is \(0\). - For \(z+2\), identify the value of \(c\) in the form \(z-c\). - After confirming the factor in part 2, divide once and factor the quadratic quotient.

1) By the remainder theorem, the remainder on division by \(x-1\) is \(P(1)\). Compute \(P(1)=1+3+4=8\). The remainder is \(8\), so \(x-1\) is not a factor. 2) Because \(z+2=z-(-2)\), evaluate \(Q(-2)=(-2)^3-6(-2)^2+32=-8-24+32=0\). By the factor theorem, \(z+2\) is a factor. Use synthetic value \(-2\) with coefficients \(1,-6,0,32\): bring down \(1\); multiply by \(-2\) and add to get \(-8\); multiply by \(-2\) and add to get \(16\); multiply by \(-2\) and add to get remainder \(0\). The quotient is \(z^2-8z+16\). Since \(z^2-8z+16=(z-4)^2\), the complete factorization is \(Q(z)=(z+2)(z-4)^2\).

1) Remainder \(8\); \(x-1\) is not a factor. 2) \(Q(-2)=0\), so \(z+2\) is a factor, and \(Q(z)=(z+2)(z-4)^2\).

- Substitute the proposed zero into the expression. - Group the first two terms and the last two terms to reveal a common binomial factor. - Consider whether any real number has a square equal to \(-1\).

1. Substitute \(x = 2\): \(T(2) = 2^3 - 2(2^2) + 2 - 2 = 8 - 8 + 2 - 2 = 0\). Therefore, \(x = 2\) is a zero. 2. Factor by grouping: \(x^3 - 2x^2 + x - 2 = x^2(x - 2) + 1(x - 2)\). 3. Factor out \(x - 2\): \(T(x) = (x - 2)(x^2 + 1)\). 4. The equation \(x^2 + 1 = 0\) would require \(x^2 = -1\), which has no real solution. Therefore, \(x^2 + 1\) has no real linear factors.

a) \(T(2) = 0\) b) \(T(x) = (x - 2)(x^2 + 1)\) c) \(x^2 + 1 > 0\) for every real \(x\), so it has no real zeros and no real linear factors.

- Represent the zeros as \(r\) and \(3r\). - Use the zeros to write the monic quadratic as a product of two linear factors. - Expand the factors and compare coefficients with the given quadratic.

1. Let the zeros be \(r\) and \(3r\). 2. Because the quadratic is monic, its factored form is \((x-r)(x-3r)\). 3. Expand: \((x-r)(x-3r)=x^2-4rx+3r^2\). 4. Compare the constant term with \(x^2+px+12\): \(3r^2=12\), so \(r^2=4\) and \(r=2\) or \(r=-2\). 5. Compare the coefficient of \(x\): \(p=-4r\). Thus \(p=-8\) when \(r=2\), and \(p=8\) when \(r=-2\).

\(p = -8\) or \(p = 8\)

- Use the given zero in the polynomial to determine \(q\). - After finding \(q\), factor the resulting quadratic. - Set each linear factor equal to zero to identify the zeros.

1. Since \(x = -1\) is a zero, \(P(-1) = 0\). 2. Substitute \(x = -1\): \((-1)^2 + 5(-1) + q = 0\). 3. Simplify: \(1 - 5 + q = 0\), so \(q = 4\). 4. Then \(P(x) = x^2 + 5x + 4 = (x + 1)(x + 4)\). 5. Therefore, the zeros are \(x = -1\) and \(x = -4\).

\(q = 4\); \(P(x) = (x + 1)(x + 4)\); zeros: \(x = -1\) and \(x = -4\)

- Use the factor theorem to connect the factor \(x + 2\) with the zero \(-2\). - Substitute the zero to determine \(b\). - Divide or compare coefficients to find the remaining factor.

1. If \(x + 2\) is a factor, then \(x = -2\) is a zero. 2. Substitute \(x = -2\): \(3(-2)^2 + b(-2) - 10 = 0\). 3. Simplify: \(12 - 2b - 10 = 0\), so \(b = 1\). 4. The expression is \(3x^2 + x - 10\). Since one factor is \(x + 2\), the other factor is \(3x - 5\). 5. Therefore, \(3x^2 + x - 10 = (x + 2)(3x - 5)\).

1. \(b = 1\) 2. \((x + 2)(3x - 5)\)

- A prescribed remainder gives a prescribed function value. - Substitute the divisor's zero into the polynomial before solving for \(k\). - Do not set the function value to \(0\) unless the problem says the divisor is a factor.

1. The remainder theorem gives \(p(2)=7\). 2. Compute \(p(2)=16+4k-6+1=4k+11\). 3. Solve \(4k+11=7\), giving \(4k=-4\) and \(k=-1\).

\(k=-1\)

- Convert the factor condition and the remainder condition into two function-value equations. - Be careful that \(x+2\) corresponds to the input \(-2\). - Solve the resulting two linear equations for the two unknown coefficients.

1. Since \(x-1\) is a factor, \(p(1)=0\), so \(1+a+b=0\). 2. The divisor \(x+2\) corresponds to \(x-(-2)\), so \(p(-2)=9\). This gives \(-8-2a+b=9\). 3. From the first equation, \(a+b=-1\). From the second, \(-2a+b=17\). 4. Subtract the first equation from the second: \(-3a=18\), so \(a=-6\). Then \(b=5\).

\(a=-6\) and \(b=5\)

- The theorem gives the remainder from one evaluation. - For the verification, use the same value \(3\) in synthetic division. - Compare the final synthetic entry with the theorem result.

a) By the remainder theorem, the remainder is \(f(3)=27-18+12+1=22\). b) Use synthetic value \(3\) with coefficients \(1,-2,4,1\). Bring down \(1\); multiply by \(3\) and add to get \(1\); multiply by \(3\) and add to get \(7\); multiply by \(3\) and add to get remainder \(22\). The quotient is \(x^2+x+7\), confirming the same remainder.

a) \(22\) b) Synthetic division also gives remainder \(22\).

- Recall exactly which part of polynomial division the remainder theorem identifies. - A quotient is a polynomial, while the theorem gives one number. - Consider what a nonzero remainder says about factor status.

1. The remainder theorem states that \(p(4)\) is the remainder when dividing by \(x-4\), not the quotient. 2. Therefore, the remainder is \(57\). 3. Because the remainder is nonzero, \(x-4\) is not a factor of \(p(x)\). The theorem alone does not determine the quotient.

The student confused the quotient with the remainder. The remainder is \(57\), so \(x-4\) is not a factor; the quotient is not determined by \(p(4)\) alone.

- Start from the polynomial division identity given in the problem. - Substitute the value that makes the divisor \(x-c\) equal to \(0\). - Connect “is a factor” with having remainder \(0\).

1. Substitute \(x=c\) into the division identity: \(p(c)=(c-c)q(c)+r=0\cdot q(c)+r=r\). 2. Thus, the function value \(p(c)\) equals the remainder. 3. The divisor \(x-c\) is a factor exactly when the remainder is \(0\). Since \(p(c)=r\), this happens exactly when \(p(c)=0\).

Substituting \(x=c\) gives \(p(c)=r\). Therefore, \(x-c\) is a factor exactly when \(r=0\), which is equivalent to \(p(c)=0\).

- Ask whether the problem needs a quotient or only a factor decision. - Each linear factor candidate determines one input value to test. - A zero function value is the deciding condition.

1. The factor theorem is more efficient because only factor status is needed. 2. Evaluate \(f(1)=3-2+1-2=0\). Therefore, \(x-1\) is a factor. 3. Evaluate \(f(-1)=-3+2-1-2=-4\ne0\). Therefore, \(x+1\) is not a factor.

Use the factor theorem. \(x-1\) is a factor, and \(x+1\) is not.

- Translate the factor condition and the prescribed remainder into two evaluation equations. - Solve for the two unknown coefficients before trying to factor the polynomial. - Use the known factor to reduce the resulting cubic after the parameters are found.

a) Since \(x-2\) is a factor, \(p(2)=0\): \(8+4a+2b+6=0\), so \(2a+b=-7\). The divisor \(x+1\) gives \(p(-1)=12\): \(-1+a-b+6=12\), so \(a-b=7\). Solving the system gives \(a=0\) and \(b=-7\). b) The polynomial is \(p(x)=x^3-7x+6\). Since \(x-2\) is a factor, divide to get \(x^2+2x-3=(x-1)(x+3)\). Therefore, \(p(x)=(x-2)(x-1)(x+3)\).

a) \(a=0\), \(b=-7\) b) \(p(x)=(x-2)(x-1)(x+3)\)