Algebra 2 Dividing Polynomials and the Remainder Theorem Test Solutions and Scoring (PDF)

1. Divide the leading terms: \(2x^{3} \div 2x\) gives \(x^{2}\). Multiply: \(x^{2}(2x – 1) = 2x^{3} – x^{2}\). Subtract: \(2x^{3} + 5x^{2} – 14x + 8 – (2x^{3} – x^{2}) = 6x^{2} – 14x + 8\).
2. Divide the leading terms: \(6x^{2} \div 2x\) gives \(3x\). Multiply: \(3x(2x – 1) = 6x^{2} – 3x\). Subtract: \(6x^{2} – 14x + 8 – (6x^{2} – 3x) = -11x + 8\).
3. Divide the leading terms: \(-11x \div 2x\) gives \(-\frac{11}{2}\). Multiply: \(-\frac{11}{2}(2x – 1) = -11x + \frac{11}{2}\). Subtract: \(-11x + 8 – (-11x + \frac{11}{2}) = \frac{5}{2}\).

Quotient \(x^{2} + 3x – \frac{11}{2}\), remainder \(\frac{5}{2}\).

Check: \((2x – 1)(x^{2} + 3x – \frac{11}{2}) + \frac{5}{2} = 2x^{3} + 5x^{2} – 14x + 8\).

\(x^{2} + 3x – \frac{11}{2} + \frac{\frac{5}{2}}{2x – 1}\)

Scoring: 1 point for the first quotient term, 1 point for the remaining steps, 1 point for the quotient and remainder.

Coefficients with a 0 for \(x\): \(3, -5, 0, 4\). Bottom row for \(k = 2\): 3, 1, 2, 8.

\(3x^{2} + x + 2 + \frac{8}{x – 2}\)

Scoring: 1 point for including the 0 coefficient, 1 point for the bottom row, 1 point for the answer form.

\(p(-3) = -27 + 18 + 27 – 18 = 0\).

Remainder 0; yes, x + 3 is a factor

Scoring: 1 point for evaluating p(-3), 1 point for the conclusion.

Zeros -1, 2, 3 give \(p(x) = (x + 1)(x – 2)(x – 3)\).

\(p(0) = (1)(-2)(-3) = 6\).

p(x) = (x + 1)(x – 2)(x – 3); p(0) = 6

Scoring: 1 point for the factored form, 1 point for p(0).

\(p(5) = 125 – 75 + 5k + 10 = 60 + 5k = 0\), so \(k = -12\).

\(k = -12\)

Scoring: 1 point for p(5) = 0, 1 point for k.

1. Divide the leading terms: \(x^{2} \div x\) gives \(x\). Multiply: \(x(x + 1) = x^{2} + x\). Subtract: \(x^{2} + 4x + 7 – (x^{2} + x) = 3x + 7\).
2. Divide the leading terms: \(3x \div x\) gives \(3\). Multiply: \(3(x + 1) = 3x + 3\). Subtract: \(3x + 7 – (3x + 3) = 4\).

Quotient \(x + 3\), remainder \(4\).

Check: \((x + 1)(x + 3) + 4 = x^{2} + 4x + 7\).

\(x + 3 + \frac{4}{x + 1}\)

Scoring: 1 point for the quotient, 1 point for the remainder term.

Width = volume divided by (length times height) = \((x^{3} + 8x^{2} + 17x + 10) \div (x^{2} + 6x + 5)\).

1. Divide the leading terms: \(x^{3} \div x^{2}\) gives \(x\). Multiply: \(x(x^{2} + 6x + 5) = x^{3} + 6x^{2} + 5x\). Subtract: \(x^{3} + 8x^{2} + 17x + 10 – (x^{3} + 6x^{2} + 5x) = 2x^{2} + 12x + 10\).
2. Divide the leading terms: \(2x^{2} \div x^{2}\) gives \(2\). Multiply: \(2(x^{2} + 6x + 5) = 2x^{2} + 12x + 10\). Subtract: \(2x^{2} + 12x + 10 – (2x^{2} + 12x + 10) = 0\).

Width x + 2 inches

Scoring: 1 point for dividing by the product of length and height, 1 point for the width.

Jamal left out the 0 coefficient for \(x^{2}\). The correct coefficients are \(1, 0, 2, -3\).

Bottom row with \(k = 1\): 1, 1, 3, 0.

Quotient \(x^{2} + x + 3\), remainder 0

Scoring: 1 point for identifying the missing zero, 1 point for the correct quotient and remainder.

1. Divide the leading terms: \(x^{3} \div x\) gives \(x^{2}\). Multiply: \(x^{2}(x + 1) = x^{3} + x^{2}\). Subtract: \(x^{3} + 6x^{2} + 11x + 6 – (x^{3} + x^{2}) = 5x^{2} + 11x + 6\).
2. Divide the leading terms: \(5x^{2} \div x\) gives \(5x\). Multiply: \(5x(x + 1) = 5x^{2} + 5x\). Subtract: \(5x^{2} + 11x + 6 – (5x^{2} + 5x) = 6x + 6\).
3. Divide the leading terms: \(6x \div x\) gives \(6\). Multiply: \(6(x + 1) = 6x + 6\). Subtract: \(6x + 6 – (6x + 6) = 0\).

Quotient \(x^{2} + 5x + 6\), remainder 0.

Check: \((x + 1)(x^{2} + 5x + 6) + 0 = x^{3} + 6x^{2} + 11x + 6\).

\(x^{2} + 5x + 6\)

\(x^{2} + 5x + 6 = (x + 2)(x + 3)\).

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

Scoring: 1 point for the quotient, 1 point for the complete factorization.