A cubic rarely factorises at a glance. Find one linear factor by trying a few values, divide it out by long division or by comparing coefficients, and you are left with a quadratic that you already know how to handle.
To factorise a cubic \(p(x)\): (1) find a linear factor by trial, using \((x - a)\) where \(a\) divides the constant term, or \((qx - a)\) where \(q\) divides the leading coefficient, and checking \(p\) at the root gives \(0\); (2) write \(p(x) = (\text{linear factor}) \times (\text{quadratic factor})\), by long division or by comparing coefficients; (3) factorise the quadratic factor, or use its discriminant to show it has no linear factors.
Fig. 1: the cubic y = x³ + x² - 16x + 20 = (x - 2)²(x + 5). Each linear factor (x - a) gives a point where the curve meets the x-axis; the repeated factor (x - 2)² makes the curve touch the axis instead of crossing it
Section 01
Find the first factor by trial.
From the factor theorem, \((x - a)\) is a factor of \(p(x)\) exactly when \(p(a) = 0\). You cannot try every number, but you do not need to: if \(p(x) = x^{3} + \dots + d\) factorises with whole numbers, \(a\) must be a factor of the constant term \(d\), because the constant terms of the factors multiply to give \(d\). When the leading coefficient is not 1, also try fractions.
Leading coefficient 1: try \(x = \pm\) each factor of the constant term, smallest first: \(\pm 1\), then \(\pm 2\), and so on.
Leading coefficient \(q \ne 1\): also try \(x = \pm\dfrac{a}{b}\), where \(a\) divides the constant term and \(b\) divides \(q\). A root \(x = \dfrac{a}{b}\) gives the factor \((bx - a)\).
Translate with care: \(p(3) = 0\) gives the factor \((x - 3)\); \(p(-3) = 0\) gives \((x + 3)\).
Paper 1 is non-calculator: keep the arithmetic tidy, one term at a time.
Worked example Factors of the constant term
Find a linear factor of \(p(x) = x^{3} - 2x^{2} - 5x + 6\).
The constant term is \(6\), so the candidates are \(x = \pm 1, \pm 2, \pm 3, \pm 6\).
\(p(1) = 1 - 2 - 5 + 6 = 0\).
So \((x - 1)\) is a factor. (Carrying on would show \(p(-2) = 0\) and \(p(3) = 0\) too, but one factor is enough to start.)
Answer: \((x - 1)\) is a factor
Worked example When the leading coefficient is not 1
Find a linear factor of \(p(x) = 2x^{3} + x^{2} + 5x - 3\).
The root \(x = \dfrac{1}{2}\) gives the factor \((2x - 1)\); multiply by \(2\) to clear the fraction.
Answer: \((2x - 1)\) is a factor
Your turn
Find a linear factor of \(x^{3} + 2x^{2} - x - 2\).
Try \(x = 1\): \(1 + 2 - 1 - 2 = 0\), so \((x - 1)\) is a factor. (\((x + 1)\) and \((x + 2)\) also work.)
Examiner trap
The sign flips between the root and the factor. If \(p(-2) = 0\) the factor is \((x + 2)\), not \((x - 2)\). And a fractional root \(x = \dfrac{1}{2}\) gives \((2x - 1)\), not \((x - 2)\).
Section 02
Long division or inspection.
Once you have one linear factor, write the cubic as (linear factor) \(\times\) (quadratic factor). There are two ways to find the quadratic. Long division works like dividing numbers: divide the leading terms, multiply back, subtract, bring down, repeat. Keep each power in its own column; the remainder must be \(0\) if you really have a factor. Below, \(x^{3} - 2x^{2} - 5x + 6\) is divided by \((x - 1)\).
\(x^{2}\)
\(-\,x\)
\(-\,6\)
\(x - 1\)
\(x^{3}\)
\(-\,2x^{2}\)
\(-\,5x\)
\(+\,6\)
\(-\)
\(x^{3}\)
\(-\,x^{2}\)
\(-\,x^{2}\)
\(-\,5x\)
\(-\)
\(-\,x^{2}\)
\(+\,x\)
\(-\,6x\)
\(+\,6\)
\(-\)
\(-\,6x\)
\(+\,6\)
\(0\)
The quotient sits on the top line: \(x^{3} - 2x^{2} - 5x + 6 = (x - 1)(x^{2} - x - 6)\). The other method, inspection, writes \((x - 1)(ax^{2} + bx + c)\) and compares coefficients: \(a\) comes from the \(x^{3}\) terms, \(c\) from the constants, and \(b\) from the \(x^{2}\) terms. The \(x\) terms then give a free check.
Worked example Long division with a missing term
Given that \((x + 1)\) is a factor of \(x^{3} - 7x - 6\), find the quadratic factor.
There is no \(x^{2}\) term, so write the dividend as \(x^{3} + 0x^{2} - 7x - 6\) to keep the columns lined up.
The quotient is \(x^{2} - x - 12\), remainder \(0\): \(x^{3} - 3x^{2} - 10x + 24 = (x - 2)(x^{2} - x - 12)\).
Examiner trap
The subtraction step is where marks go. Subtract the WHOLE product: \(-2x^{2} - (-x^{2}) = -x^{2}\), not \(-3x^{2}\). If you do not get remainder \(0\) after dividing by a factor, an earlier line is wrong; go back rather than carrying a remainder.
Stage 1 · Learn
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Which of these is a factor of \(x^{3} - 4x^{2} + x + 6\)?
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You want a linear factor of \(3x^{3} + 5x^{2} - 4x + 2\) by trial. Which list contains every root \(x\) worth testing?
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What is the quotient when \(x^{3} + 2x^{2} - 5x - 6\) is divided by \((x + 1)\)?
'Factorise completely' means you must also deal with the quadratic factor. Factorise it if you can. If you cannot spot factors, test its discriminant: \(b^{2} - 4ac < 0\) means it has no real roots, so no linear factors, and the answer stops at (linear) \(\times\) (quadratic). When the quadratic factor contains the linear factor again, the cubic has a repeated factor (Fig. 1).
Stopping at \((x - 1)(x^{2} - x - 6)\) when the question says 'completely' loses the last mark: \(x^{2} - x - 6 = (x - 3)(x + 2)\). And \(x^{2} + 4\) is NOT \((x + 2)^{2}\) or \((x + 2)(x - 2)\); check by expanding.
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