IGCSE · Practice questions · Factorising polynomials

One root opens the whole cubic.

Six original Cambridge-style questions on factorising polynomials: finding a first factor by trial, long division, comparing coefficients, repeated factors and quadratic factors that cannot be factorised.

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

Trial, divide, then finish the quadratic.

Trial: test \(x = \pm\) factors of the constant term (and \(\pm\dfrac{a}{b}\) if the leading coefficient is not 1) until \(p(x) = 0\).
Translate: \(p(a) = 0\) gives \((x - a)\); \(p\left(\dfrac{a}{b}\right) = 0\) gives \((bx - a)\).
Divide by long division (write \(0x^{2}\) for a missing term) or compare coefficients; remainder \(0\) checks the factor.
Finish: factorise the quadratic factor, or show \(b^{2} - 4ac < 0\).
Tidy: write a repeated factor as a square, and check by expanding if time allows.

Recall

[2 marks]

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

\(x^{2}(x + 2) = x^{3} + 2x^{2}\), leaving \(2x^{2} + x\); \(2x(x + 2) = 2x^{2} + 4x\), leaving \(-3x - 6\); \(-3(x + 2) = -3x - 6\), leaving \(0\). ✓
The quotient is \(x^{2} + 2x - 3\). ✓

Calculation

[4 marks]

Factorise completely \(x^{3} + 2x^{2} - 11x - 12\).

\(p(1) = 1 + 2 - 11 - 12 = -20\); \(p(-1) = -1 + 2 + 11 - 12 = 0\), so \((x + 1)\) is a factor. ✓
\((x + 1)(x^{2} + bx - 12)\): the \(x^{2}\) terms give \(b + 1 = 2\), so \(b = 1\). Check the \(x\) terms: \(-12 + b = -11\). ✓
\(x^{2} + x - 12 = (x + 4)(x - 3)\). ✓
\(x^{3} + 2x^{2} - 11x - 12 = (x + 1)(x + 4)(x - 3)\). ✓

Calculation

[4 marks]

Factorise completely \(2x^{3} + 3x^{2} - 11x - 6\).

\(p(1) = -12\), \(p(-1) = 6\), \(p(2) = 16 + 12 - 22 - 6 = 0\), so \((x - 2)\) is a factor. ✓
\((x - 2)(2x^{2} + bx + 3)\): the \(x^{2}\) terms give \(b - 4 = 3\), so \(b = 7\). Check the \(x\) terms: \(3 - 2b = -11\). ✓
\(2x^{2} + 7x + 3 = (2x + 1)(x + 3)\). ✓
\(2x^{3} + 3x^{2} - 11x - 6 = (x - 2)(2x + 1)(x + 3)\). ✓

Analysis

[4 marks]

\(p(x) = 3x^{3} - 2x^{2} + 6x - 4\).

(a) Show that \((3x - 2)\) is a factor of \(p(x)\).

(b) Write \(p(x)\) as the product of a linear factor and a quadratic factor, and show that \(p(x)\) has no other linear factors.

(a) \(p\left(\dfrac{2}{3}\right) = 3\left(\dfrac{8}{27}\right) - 2\left(\dfrac{4}{9}\right) + 4 - 4 = \dfrac{8}{9} - \dfrac{8}{9} = 0\), so \((3x - 2)\) is a factor. ✓
(b) \((3x - 2)(x^{2} + bx + 2)\): the \(x^{2}\) terms give \(3b - 2 = -2\), so \(b = 0\). Check the \(x\) terms: \(6 - 2b = 6\). ✓
\(p(x) = (3x - 2)(x^{2} + 2)\). ✓
For \(x^{2} + 2\): \(b^{2} - 4ac = 0 - 8 = -8 < 0\), so it has no real roots and \(p(x)\) has no other linear factors. ✓

Analysis

[4 marks]

Factorise completely \(4x^{3} - 12x^{2} + 9x - 2\), and state which factor is repeated.

\(p(1) = 4 - 12 + 9 - 2 = -1\); \(p(2) = 32 - 48 + 18 - 2 = 0\), so \((x - 2)\) is a factor. ✓
\((x - 2)(4x^{2} + bx + 1)\): the \(x^{2}\) terms give \(b - 8 = -12\), so \(b = -4\). Check the \(x\) terms: \(1 - 2b = 9\). ✓
\(4x^{2} - 4x + 1 = (2x - 1)^{2}\). ✓
\(4x^{3} - 12x^{2} + 9x - 2 = (x - 2)(2x - 1)^{2}\); the factor \((2x - 1)\) is repeated. ✓

Proof

[5 marks]

It is given that \(2x^{3} + x^{2} + kx + 6 \equiv (x + 2)(2x^{2} + ax + b)\) for all values of \(x\).

(a) Find the values of the constants \(a\), \(b\) and \(k\).

(b) Show that \(2x^{3} + x^{2} + kx + 6\) cannot be factorised into three linear factors.

(a) Expand: \((x + 2)(2x^{2} + ax + b) = 2x^{3} + (a + 4)x^{2} + (b + 2a)x + 2b\). ✓
\(x^{2}\) terms: \(a + 4 = 1\), so \(a = -3\). Constants: \(2b = 6\), so \(b = 3\). ✓
\(x\) terms: \(k = b + 2a = 3 - 6 = -3\). ✓
(b) The quadratic factor is \(2x^{2} - 3x + 3\), with \(b^{2} - 4ac = 9 - 24 = -15 < 0\). ✓
It has no real roots, so no linear factors: the cubic is \((x + 2)(2x^{2} - 3x + 3)\) and cannot be written as three linear factors. ✓

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.