Precalculus Polynomial Division and the Factor Theorem Worksheet Solutions (PDF)

Precalculus math worksheet solutions: Polynomial Division and the Factor Theorem – Maths PDF

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This precalculus polynomial division answers for Precalculus covers polynomial division and the factor theorem (Common Core A-APR.B.2, A-APR.D.6, A-APR.B.3), with a free PDF to print.

Use this key to check each division and zero. The synthetic division tables show every step, and final answers are in bold.

Precalculus · Common Core State Standards: A-APR.B.2, A-APR.D.6, A-APR.B.3

Precalculus polynomial division answers: exercises 1 to 10

Exercise 1 – Divide a quadratic

\(x^{2}\div x=x\); subtract \(x^{2}+3x\) to get \(4x+12\); \(4x\div x=4\); subtract \(4x+12\) to get 0.

\(x+4\) with remainder \(0\)

Exercise 2 – Long division, leading coefficient 2

\(2x^{2}\div x=2x\); subtract \(2x^{2}+6x\) to get \(-x-3\); \(-x\div x=-1\); subtract \(-x-3\) to get 0.

\(2x-1\)

Exercise 3 – Remainder from long division

Coefficients 1, 2, -4, 7 with c = 1.

The bottom row is \(1,\ 3,\ -1\) with remainder \(6\).

\(x^{2}+3x-1+\frac{6}{x-1}\)

Exercise 4 – Synthetic division

c = 2.

The bottom row is \(1,\ -2,\ -3,\ 0\).

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

Exercise 5 – Synthetic division with c = -1

c = -1.

The bottom row is \(2,\ -5,\ 10,\ -11\).

\(2x^{2}-5x+10-\frac{11}{x+1}\)

Exercise 6 – Remainder Theorem

The remainder is \(f(2)=16-12+5\).

Exercise 7 – Is it a factor?

\(f(3)=27-27=0\), so yes. Division gives \(x^{2}+3x+9\), which does not factor over the integers (discriminant \(9-36< 0\)).

\(x^{3}-27=(x-3)(x^{2}+3x+9)\)

Exercise 8 – Test a factor

\(f(-2)=-8+12+8-12=0\).

Yes, because the remainder is 0

Exercise 9 – Find an unknown coefficient

\(f(2)=8+4k-8-4=4k-4=0\), so \(k=1\).

\(k=1\)

Exercise 10 – Factor completely

Synthetic division with \(c=1\) on \(1,0,-7,6\) gives \(1,1,-6\) with remainder 0, so \(x^{3}-7x+6=(x-1)(x^{2}+x-6)=(x-1)(x+3)(x-2)\).

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

Exercises 11 and up

Exercise 11 – Rational root candidates

\(p\in\{\pm1,\pm2,\pm3,\pm6\}\) and \(q\in\{\pm1,\pm2\}\). Dividing gives twelve candidates.

\(\pm1,\ \pm2,\ \pm3,\ \pm6,\ \pm\frac12,\ \pm\frac32\)

Exercise 12 – Find all zeros

c = 3 leaves 2, 3, -2 and remainder 0.

The quotient is \(2x^{2}+3x-2=(2x-1)(x+2)\).

\(x=3,\ \frac12,\ -2\)

Exercise 13 – Missing terms

Placeholders 0 for x^2 and x.

The bottom row is \(1,\ 2,\ 4,\ 0\).

\(x^{2}+2x+4\), remainder \(0\)

Exercise 14 – Biquadratic

Let \(u=x^{2}\): \(u^{2}-5u+4=(u-1)(u-4)=0\), so \(x^{2}=1\) or \(x^{2}=4\).

\(x=\pm1,\ \pm2\)

Exercise 15 – Quartic zeros

Divide by \(x-1\): \(x^{3}-7x-6\). Divide by \(x+1\): \(x^{2}-x-6=(x-3)(x+2)\).

\(x=1,\ -1,\ 3,\ -2\)

Exercise 16 – Divide by a non-monic linear divisor

\(6x^{2}\div 2x=3x\); subtract \(6x^{2}+9x\) to get \(-8x-5\); \(-8x\div 2x=-4\); subtract \(-8x-12\) to get \(7\).

\(3x-4+\frac{7}{2x+3}\)

Exercise 17 – Box dimensions

Dimensions: x + 2, x + 3, x + 4.

Synthetic division with \(c=-2\): \(1,9,26,24\to 1,7,12\), remainder 0. The quotient \(x^{2}+7x+12=(x+3)(x+4)\).

\(x+3\) and \(x+4\) feet

Exercise 18 – Break-even years

Zeros at t = 3 and t = 4.

\(P(4)=64-80-8+24=0\). Division by \(t-4\) gives \(t^{2}-t-6=(t-3)(t+2)\). The zeros are \(4,\ 3,\ -2\); only \(t\ge 0\) counts.

\(t=3\) and \(t=4\) years

Exercise 19 – Multiplicity of a zero

\(f\left(\frac12\right)=\frac12-2+\frac52-1=0\). Factoring gives \((2x-1)^{2}(x-1)\).

\(\frac12\) (multiplicity 2) and \(1\) (multiplicity 1)

Exercise 20 – Polynomial with integer coefficients

\(f(x)=(x-2)(2x+1)(x-3)\). First \((x-2)(x-3)=x^{2}-5x+6\), then multiply by \((2x+1)\): \(2x^{3}-9x^{2}+7x+6\).

\(2x^{3}-9x^{2}+7x+6\)

Key facts behind this precalculus polynomial division answers

Key facts

f(x) = d(x)q(x) + r(x), with r of lower degree than d.
Remainder Theorem: dividing by x – c leaves f(c).
Factor Theorem: x – c is a factor if and only if f(c) = 0.
Rational Root Theorem: p divides the constant term and q divides the leading coefficient.
Use 0 as a placeholder coefficient for each missing power.

How to use this precalculus 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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