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

Algebra 2 math worksheet solutions: Dividing Polynomials and the Remainder Theorem – Maths PDF

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This algebra 2 polynomial division answers for Algebra 2 covers dividing polynomials and the remainder theorem (Common Core A-APR.B.2, A-APR.D.6), with a free PDF to print.

Every worksheet problem on polynomial division and the Remainder Theorem is solved here, one step at a time. Compare your method as well as your final answer, because the steps show where a missing zero coefficient can go wrong. Final answers appear in bold so you can find them quickly when you only need a check.

Algebra 2 · Common Core State Standards: A-APR.B.2, A-APR.D.6

Algebra 2 polynomial division answers: exercises 1 to 10

Exercise 1 – Divide, no remainder

Divide the leading terms: \(x^{2} \div x\) gives \(x\). Multiply: \(x(x + 3) = x^{2} + 3x\). Subtract: \(x^{2} + 7x + 12 – (x^{2} + 3x) = 4x + 12\).
Divide the leading terms: \(4x \div x\) gives \(4\). Multiply: \(4(x + 3) = 4x + 12\). Subtract: \(4x + 12 – (4x + 12) = 0\).

Quotient \(x + 4\), remainder 0.

Check: \((x + 3)(x + 4) + 0 = x^{2} + 7x + 12\).

\(x + 4\)

Exercise 2 – Divide, remainder

Divide the leading terms: \(x^{2} \div x\) gives \(x\). Multiply: \(x(x – 2) = x^{2} – 2x\). Subtract: \(x^{2} – 5x + 7 – (x^{2} – 2x) = -3x + 7\).
Divide the leading terms: \(-3x \div x\) gives \(-3\). Multiply: \(-3(x – 2) = -3x + 6\). Subtract: \(-3x + 7 – (-3x + 6) = 1\).

Quotient \(x – 3\), remainder \(1\).

Check: \((x – 2)(x – 3) + 1 = x^{2} – 5x + 7\).

\(x – 3 + \frac{1}{x – 2}\)

Exercise 3 – Leading coefficient 2

Divide the leading terms: \(10x^{2} \div 2x\) gives \(5x\). Multiply: \(5x(2x + 3) = 10x^{2} + 15x\). Subtract: \(10x^{2} – x – 24 – (10x^{2} + 15x) = -16x – 24\).
Divide the leading terms: \(-16x \div 2x\) gives \(-8\). Multiply: \(-8(2x + 3) = -16x – 24\). Subtract: \(-16x – 24 – (-16x – 24) = 0\).

Quotient \(5x – 8\), remainder 0.

Check: \((2x + 3)(5x – 8) + 0 = 10x^{2} – x – 24\).

\(5x – 8\)

Exercise 4 – Cubic long division

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

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

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

\(x^{2} + x – 6\)

Exercise 5 – Synthetic division

Synthetic division by x – 2: the last number is the remainder.

Bring down 2. Multiply by 2 and add, column by column: 2, 1, 3, 1.

Quotient \(2x^{2} + x + 3\), remainder \(1\).

\(2x^{2} + x + 3 + \frac{1}{x – 2}\)

Exercise 6 – Missing term

Write the coefficients with a 0 for \(x^{2}\): \(1, 0, -7, 6\).

Bottom row with \(k = 2\): 1, 2, -3, 0.

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

Exercise 7 – Divisor x + 3

Coefficients with a zero for \(x^{2}\): \(1, 3, 0, -1, 4\).

Bottom row: 1, 0, 0, -1, 7.

\(x^{3} – 1 + \frac{7}{x + 3}\)

Exercise 8 – Remainder Theorem, zero

By the Remainder Theorem, the remainder is \(p(3) = 27 – 36 + 3 + 6\).

\(p(3) = 0\): remainder 0

Exercise 9 – Remainder Theorem

\(x + 2 = x – (-2)\), so evaluate \(p(-2) = 2(-8) + 4 + 10 + 4\).

Remainder 2

Exercise 10 – Factor check

\(p(4) = 64 – 80 + 8 + 8 = 0\).

The remainder is 0.

Yes, x – 4 is a factor

Exercises 11 and up

Exercise 11 – Find k

Since \(x + 1 = x – (-1)\), set \(p(-1) = -1 + k + 5 + 3 = k + 7\) equal to 0.

\(k = -7\)

Exercise 12 – Read the graph

Zeros -2, 1, 3.

Each zero \(k\) gives a factor \(x – k\): \(x + 2\), \(x – 1\), \(x – 3\).

Check the leading coefficient: \((x + 2)(x – 1)(x – 3)\) has leading coefficient 1, matching \(p(x)\).

\(p(x) = (x + 2)(x – 1)(x – 3)\)

Exercise 13 – Factor completely

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

\(x^{2} – 5x + 6\)

Factor the quotient: \(x^{2} – 5x + 6 = (x – 2)(x – 3)\).

\((x – 1)(x – 2)(x – 3)\)

Exercise 14 – Rewrite a rational expression

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

Quotient \(x + 1\), remainder \(3\).

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

\(x + 1 + \frac{3}{x + 2}\)

Exercise 15 – Rewrite with a cubic

Divide the leading terms: \(3x^{3} \div x\) gives \(3x^{2}\). Multiply: \(3x^{2}(x – 1) = 3x^{3} – 3x^{2}\). Subtract: \(3x^{3} – 2x^{2} + 4 – (3x^{3} – 3x^{2}) = x^{2} + 4\).
Divide the leading terms: \(x^{2} \div x\) gives \(x\). Multiply: \(x(x – 1) = x^{2} – x\). Subtract: \(x^{2} + 4 – (x^{2} – x) = x + 4\).
Divide the leading terms: \(x \div x\) gives \(1\). Multiply: \(1(x – 1) = x – 1\). Subtract: \(x + 4 – (x – 1) = 5\).

Quotient \(3x^{2} + x + 1\), remainder \(5\).

Check: \((x – 1)(3x^{2} + x + 1) + 5 = 3x^{3} – 2x^{2} + 4\).

\(3x^{2} + x + 1 + \frac{5}{x – 1}\)

Exercise 16 – Box dimensions

Height found: x + 4.

Height = volume divided by length times width: \((x^{3} + 9x^{2} + 26x + 24) \div (x^{2} + 5x + 6)\).

Divide the leading terms: \(x^{3} \div x^{2}\) gives \(x\). Multiply: \(x(x^{2} + 5x + 6) = x^{3} + 5x^{2} + 6x\). Subtract: \(x^{3} + 9x^{2} + 26x + 24 – (x^{3} + 5x^{2} + 6x) = 4x^{2} + 20x + 24\).
Divide the leading terms: \(4x^{2} \div x^{2}\) gives \(4\). Multiply: \(4(x^{2} + 5x + 6) = 4x^{2} + 20x + 24\). Subtract: \(4x^{2} + 20x + 24 – (4x^{2} + 20x + 24) = 0\).

Height \(x + 4\) inches

Exercise 17 – Rectangle length

Divide the leading terms: \(x^{3} \div x\) gives \(x^{2}\). Multiply: \(x^{2}(x + 3) = x^{3} + 3x^{2}\). Subtract: \(x^{3} + 3x^{2} – 4x – 12 – (x^{3} + 3x^{2}) = -4x – 12\).
Divide the leading terms: \(-4x \div x\) gives \(-4\). Multiply: \(-4(x + 3) = -4x – 12\). Subtract: \(-4x – 12 – (-4x – 12) = 0\).

Quotient \(x^{2} – 4\), remainder 0.

Check: \((x + 3)(x^{2} – 4) + 0 = x^{3} + 3x^{2} – 4x – 12\).

\(x^{2} – 4\)

Exercise 18 – Evaluate with synthetic division

Last entry: remainder = P(4) = 14.

Bottom row with \(k = 4\): 1, -2, 3, 14. By the Remainder Theorem \(P(4) = 14\).

1,400 individuals at t = 4

Exercise 19 – Difference of cubes

\(p(5) = 125 – 125 = 0\), so \(x – 5\) is a factor.

Divide: \(x^{3} – 125 = (x – 5)(x^{2} + 5x + 25)\).

The quadratic has discriminant \(25 – 100 = -75 < 0\), so it does not factor over the real numbers.

\((x – 5)(x^{2} + 5x + 25)\)

Exercise 20 – No division needed

\(p(1) = 1 – 1 = 0\). \(p(-1) = (-1)^{10} – 1 = 1 – 1 = 0\).

Both remainders are 0

Key facts behind this algebra 2 polynomial division answers

Key facts

a(x) / b(x) = q(x) + r(x) / b(x) with deg r < deg b.
Synthetic division works for divisors x – k.
Remainder Theorem: p(x) ÷ (x – k) has remainder p(k).
Factor Theorem: x – k is a factor if and only if p(k) = 0.
Include 0 coefficients for missing powers.

How to use this algebra 2 polynomial division answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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