Use the remainder theorem to find the remainder on division by a linear factor and the factor theorem to identify and extract factors of a polynomial

What this dot point is asking

SEAB wants you to divide one polynomial by a linear expression, to find the remainder quickly using the remainder theorem, and to use the factor theorem to test for and extract linear factors so that a cubic can be fully factorised. These two theorems turn an otherwise long division into a single substitution.

The answer

Polynomial division

Dividing a polynomial $f(x)$ by a divisor $g(x)$ gives a quotient $q(x)$ and a remainder $r(x)$ whose degree is below that of $g(x)$:

When the divisor is linear, the remainder is just a number.

The remainder theorem

Instead of dividing, substitute. When $f(x)$ is divided by $(x - a)$, the remainder equals $f(a)$:

For a divisor $(bx - a)$, the remainder is $f\!\left(\dfrac{a}{b}\right)$, the value that makes the divisor zero.

The factor theorem

The factor theorem is the special case where the remainder is zero. If $f(a) = 0$, then $(x - a)$ divides $f(x)$ exactly, so $(x - a)$ is a factor:

Factorising a cubic

To factorise a cubic, find one root by trial (test small integer factors of the constant term), use the factor theorem to confirm a linear factor, then divide to get a quadratic, which you factorise by the usual methods.

A divisor of the form bx minus a

For a divisor such as $(2x - 1)$, the value that makes it zero is $x = \tfrac{1}{2}$, so the remainder is $f\!\left(\tfrac{1}{2}\right)$ and $(2x - 1)$ is a factor exactly when $f\!\left(\tfrac{1}{2}\right) = 0$. The principle is unchanged; only the test value differs.

Two conditions, two unknowns

A common Paper 2 question gives two facts, such as two remainders, or one factor and one remainder, for a polynomial with two unknown coefficients. Each fact gives an equation in the unknowns through $f$ evaluated at a value; solving the pair of simultaneous equations pins down both coefficients.

Examples in context

Example 1. Finding unknown coefficients. A question may give two pieces of remainder or factor information and ask for two unknown constants. Each condition becomes an equation in the unknowns via $f(a)$, and solving the pair pins them down, a standard Paper 2 structured problem.

Example 2. Sketching a cubic. Once $f(x) = (x - 2)(2x + 1)(x + 3)$ is factorised, the roots $x = 2, -\tfrac{1}{2}, -3$ are exactly the $x$-intercepts of the curve $y = f(x)$, linking algebra to the shape of the graph.

Try this

Q1. Find the remainder when $x^3 + 2x^2 - x + 5$ is divided by $(x - 1)$. [2 marks]

• Cue. Remainder $= f(1) = 1 + 2 - 1 + 5 = 7$.

Q2. Show that $(x - 3)$ is a factor of $x^3 - 4x^2 + x + 6$. [2 marks]

• Cue. $f(3) = 27 - 36 + 3 + 6 = 0$, so by the factor theorem $(x - 3)$ is a factor.

Q3. Given that $(x + 2)$ is a factor of $x^3 + kx - 2$, find $k$. [3 marks]

• Cue. $f(-2) = -8 - 2k - 2 = 0$, so $-2k = 10$ and $k = -5$.