Factor Theorem

The factor theorem is the remainder theorem at zero: $(x - a)$ is a factor of $p(x)$ if and only if $p(a) = 0$. A factor divides exactly; exact division means zero remainder; zero remainder means $p(a) = 0$. Three uses cover every exam appearance.

Use 1. “Show that $(x - 3)$ is a factor”

$p(x) = x^3 - x^2 - 7x + 3$: $p(3) = 27 - 9 - 21 + 3 = 0$ ”$p(3) = 0$, therefore $(x - 3)$ is a factor.”

The computation is half the marks; the conclusion sentence is the other half. This is a show-that command: the destination is printed, so the visible route, substitution shown term by term, then the stated inference, is what’s being marked.

Use 2. Find the unknown, given a factor

$(x + 2)$ is a factor of $p(x) = 2x^3 + kx^2 - 5x - 6$. Find $k$. $p(-2) = 0$: $-16 + 4k + 10 - 6 = 0 \to 4k = 12 \to$ $k = 3$

One condition, one equation. With two unknowns, you’ll be given two facts, often one factor plus one remainder, yielding simultaneous equations. Translate each fact separately and label which is which; the translation lines are the M marks.

Use 3. Find the first factor of a cubic by trial

To factorise a cubic with no factor given, hunt a root among the factors of the constant term: for $p(x) = x^3 - 4x^2 + x + 6$, test $x = \pm 1, \pm 2, \pm 3, \pm 6$. $p(-1) = -1 - 4 - 1 + 6 = 0$ ✓ $\to (x + 1)$ is a factor, and the full factorising routine takes over. Trial values chosen from the constant’s factors, not at random, is both faster and visibly methodical.

The sign discipline (again, because it pays)

Factor $(x + 2) \to$ substitute $-2$. Factor $(2x - 1) \to$ substitute $+\frac{1}{2}$. Set the factor to zero on paper before substituting; the two seconds spent prevent the most common error in the whole topic.

Common mistakes

• The conclusion sentence omitted on “show that” parts, computation alone drops a mark
• Wrong-sign substitution
• “Is a factor” translated as $p(a) =$ remainder instead of $p(a) = 0$
• Trial roots picked from thin air instead of the constant term’s factors
• Stopping at the first factor when the question asked for a full factorisation

Full topic context: Factors of Polynomials notes · next step: factorising & solving cubics.