AS & A Level · Practice questions · Polynomial division, factor and remainder theorems

Every remainder counts.

Papers 2 and 3. Six original Cambridge-style questions on dividing polynomials and on the factor and remainder theorems, ending with a two-unknown problem solved and factorised.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Substitute the root of the divisor.

Divide: leading term, multiply back, subtract, repeat; include zero terms for missing powers.
Stop: when the remainder's degree is below the divisor's degree.
Remainder theorem: remainder on dividing by \((ax + b)\) is \(p\left(-\dfrac{b}{a}\right)\).
Factor theorem: \((ax + b)\) is a factor if and only if \(p\left(-\dfrac{b}{a}\right) = 0\).
Unknowns: one equation per condition, then solve simultaneously.

Recall

[2 marks]

Find the remainder when \(2x^3 - 3x^2 + x - 5\) is divided by \((x + 2)\).

\(p(-2) = 2(-8) - 3(4) + (-2) - 5 = -16 - 12 - 2 - 5 = -35\). ✓
Remainder \(-35\). ✓

Calculation

[3 marks]

Find the quotient and remainder when \(x^3 + 2x^2 - 5x + 1\) is divided by \((x - 3)\).

\(x^2(x - 3) = x^3 - 3x^2\); subtract: \(5x^2 - 5x\). ✓
\(5x(x - 3) = 5x^2 - 15x\); subtract: \(10x + 1\). ✓
\(10(x - 3) = 10x - 30\); subtract: \(31\). ✓
Quotient \(x^2 + 5x + 10\), remainder \(31\). (Check: \(p(3) = 27 + 18 - 15 + 1 = 31\).) ✓

Calculation

[4 marks]

Find the quotient and remainder when \(2x^4 - x^3 + 3x^2 + 4\) is divided by \(x^2 - x + 2\).

Write \(2x^4 - x^3 + 3x^2 + 0x + 4\). ✓
\(2x^2(x^2 - x + 2) = 2x^4 - 2x^3 + 4x^2\); subtract: \(x^3 - x^2 + 0x + 4\). ✓
\(x(x^2 - x + 2) = x^3 - x^2 + 2x\); subtract: \(-2x + 4\). ✓
Quotient \(2x^2 + x\), remainder \(-2x + 4\). ✓

Analysis

[5 marks]

(a) Show that \((3x + 1)\) is a factor of \(p(x) = 3x^3 - 2x^2 - 7x - 2\).

(b) Hence factorise \(p(x)\) completely and solve \(p(x) = 0\).

(a) \(p\left(-\dfrac{1}{3}\right) = 3\left(-\dfrac{1}{27}\right) - 2\left(\dfrac{1}{9}\right) + \dfrac{7}{3} - 2 = -\dfrac{1}{9} - \dfrac{2}{9} + \dfrac{21}{9} - \dfrac{18}{9} = 0\), so \((3x + 1)\) is a factor. ✓
(b) \(p(x) = (3x + 1)(x^2 - x - 2)\) (check: \(x^2\) terms \(-3x^2 + x^2 = -2x^2\); \(x\) terms \(-6x - x = -7x\)). ✓
\(x^2 - x - 2 = (x - 2)(x + 1)\), so \(p(x) = (3x + 1)(x - 2)(x + 1)\). ✓
\(x = -\dfrac{1}{3},\ 2\) or \(-1\). ✓

Analysis

[4 marks]

When \(x^3 + ax^2 - 4x + b\) is divided by \((x - 1)\) the remainder is \(-6\), and when it is divided by \((x + 2)\) the remainder is \(12\). Find \(a\) and \(b\).

\(p(1) = 1 + a - 4 + b = -6\), so \(a + b = -3\). ✓
\(p(-2) = -8 + 4a + 8 + b = 12\), so \(4a + b = 12\). ✓
Subtract: \(3a = 15\), so \(a = 5\) and \(b = -8\). ✓

Analysis

[7 marks]

The polynomial \(p(x) = 2x^3 + ax^2 + bx + 10\) has \((x + 2)\) as a factor. When \(p(x)\) is divided by \((x - 1)\) the remainder is \(-6\).

(a) Find the values of \(a\) and \(b\).

(b) Factorise \(p(x)\) completely.

(a) \(p(-2) = -16 + 4a - 2b + 10 = 0\), so \(2a - b = 3\). ✓
\(p(1) = 2 + a + b + 10 = -6\), so \(a + b = -18\). ✓
Add: \(3a = -15\), so \(a = -5\) and \(b = -13\). ✓
(b) \(p(x) = 2x^3 - 5x^2 - 13x + 10 = (x + 2)(2x^2 - 9x + 5)\). Check: \(x^2\) terms \(-9x^2 + 4x^2 = -5x^2\); \(x\) terms \(5x - 18x = -13x\). ✓
\(2x^2 - 9x + 5\) has discriminant \(81 - 40 = 41\), so it does not factorise over the integers: \(p(x) = (x + 2)(2x^2 - 9x + 5)\). ✓
(c) Put \(x = y^2\): \(y^2 = -2\) (no real solutions) or \(y^2 = \dfrac{9 \pm \sqrt{41}}{4}\). ✓
Both values \(\dfrac{9 \pm \sqrt{41}}{4}\) are positive (\(\sqrt{41} < 9\)), so \(y = \pm\dfrac{\sqrt{9 + \sqrt{41}}}{2}\) or \(y = \pm\dfrac{\sqrt{9 - \sqrt{41}}}{2}\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.