IGCSE · Practice questions · The remainder and factor theorems

Substitute, then solve.

Six original Cambridge-style questions on the remainder and factor theorems: remainders by substitution, showing that a linear expression is a factor, and finding unknown coefficients from given factors and remainders.

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

The method

Make the divisor zero, then substitute.

Find the value of \(x\) that makes the divisor zero: \(x - a\) gives \(a\), \(x + a\) gives \(-a\), \(ax - b\) gives \(\frac{b}{a}\).
Substitute into \(p(x)\), with negative values and fractions in brackets.
Read the result: zero means a factor; any other value is the remainder.
Unknown coefficients: each factor or remainder fact gives one equation; two facts give simultaneous equations.
Check your values in the original polynomial and finish 'show that' answers with a sentence.

Recall

[2 marks]

Find the remainder when \(2x^{3} + x^{2} - 5x + 4\) is divided by \(x - 2\).

\(x - 2 = 0\) when \(x = 2\). ✓
\(p(2) = 2(8) + 4 - 10 + 4 = 16 + 4 - 10 + 4 = 14\). The remainder is \(14\). ✓

Calculation

[2 marks]

Show that \(x + 2\) is a factor of \(x^{3} - 3x^{2} - 6x + 8\).

\(x + 2 = 0\) when \(x = -2\). ✓
\(p(-2) = -8 - 3(4) - 6(-2) + 8 = -8 - 12 + 12 + 8 = 0\), so \(x + 2\) is a factor. ✓

Calculation

[3 marks]

Find the remainder when \(4x^{3} - 6x^{2} + 5\) is divided by \(2x - 3\).

\(2x - 3 = 0\) when \(x = \dfrac{3}{2}\). ✓
\(p\left(\dfrac{3}{2}\right) = 4\left(\dfrac{27}{8}\right) - 6\left(\dfrac{9}{4}\right) + 5 = \dfrac{27}{2} - \dfrac{27}{2} + 5\). ✓
The remainder is \(5\). ✓

Calculation

[3 marks]

Given that \(3x + 2\) is a factor of \(3x^{3} + kx^{2} - 4\), find the value of the constant \(k\).

\(3x + 2 = 0\) when \(x = -\dfrac{2}{3}\), so \(p\left(-\dfrac{2}{3}\right) = 0\). ✓
\(3\left(-\dfrac{8}{27}\right) + k\left(\dfrac{4}{9}\right) - 4 = 0\), so \(-\dfrac{8}{9} + \dfrac{4k}{9} - 4 = 0\). ✓
Multiply by 9: \(-8 + 4k - 36 = 0\), so \(4k = 44\) and \(k = 11\). ✓

Analysis

[6 marks]

The polynomial \(p(x) = 2x^{3} + ax^{2} + bx - 6\) has a factor \(x - 2\). When \(p(x)\) is divided by \(x + 1\) the remainder is \(-12\).

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

(b) Using your values of \(a\) and \(b\), find the remainder when \(p(x)\) is divided by \(2x - 1\).

(a) \(p(2) = 16 + 4a + 2b - 6 = 0\), so \(4a + 2b = -10\), that is \(2a + b = -5\). ✓
\(p(-1) = -2 + a - b - 6 = -12\), so \(a - b = -4\). ✓
Add: \(3a = -9\), so \(a = -3\); then \(b = -5 - 2(-3) = 1\). ✓
Check with \(p(x) = 2x^{3} - 3x^{2} + x - 6\): \(p(2) = 16 - 12 + 2 - 6 = 0\) and \(p(-1) = -2 - 3 - 1 - 6 = -12\). ✓
(b) \(p\left(\dfrac{1}{2}\right) = 2\left(\dfrac{1}{8}\right) - 3\left(\dfrac{1}{4}\right) + \dfrac{1}{2} - 6 = \dfrac{1}{4} - \dfrac{3}{4} + \dfrac{1}{2} - 6 = -6\). ✓
The remainder is \(-6\). ✓

Proof

[4 marks]

The polynomial \(p(x) = x^{3} + kx^{2} + (2k - 1)x + 6\), where \(k\) is a constant.

(a) Show that \(x + 2\) is a factor of \(p(x)\) for every value of \(k\).

(b) Find the value of \(k\) for which \(x - 1\) is also a factor of \(p(x)\).

(a) \(p(-2) = -8 + 4k - 2(2k - 1) + 6 = -8 + 4k - 4k + 2 + 6\). ✓
\(= 0\) whatever the value of \(k\), so \(x + 2\) is a factor for every \(k\). ✓
(b) \(p(1) = 1 + k + 2k - 1 + 6 = 3k + 6 = 0\), so \(k = -2\). ✓
(Then \(p(x) = x^{3} - 2x^{2} - 5x + 6\), the cubic in Fig. 1, which also has \(x - 3\) as a factor.) ✓

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.