Work with polynomials: apply the factor theorem and the remainder theorem, factorise cubics, and solve polynomial equations of degree three.

What this dot point is asking

A polynomial is an expression with whole-number powers of $x$, and CCEA GCSE Further Mathematics focuses on cubics: expressions of degree three. You must use the factor theorem to spot a linear factor, the remainder theorem to find what is left when you divide, and then factorise a cubic fully so you can solve it. This topic is a signature Further Mathematics skill, because it lets you handle equations that ordinary GCSE cannot.

The factor theorem

The factor theorem turns factorising into substitution. If putting $x = a$ into $f(x)$ gives zero, then $(x - a)$ divides $f(x)$ exactly. To find a first factor of a cubic, test the factors of the constant term: for $x^3 + 2x^2 - 5x - 6$, try $x = 1, -1, 2, -2, 3, -3$ until one gives zero.

Watch the sign carefully: the factor $(x + 2)$ corresponds to testing $x = -2$, because $(x + 2) = (x - (-2))$.

The remainder theorem

The remainder theorem is the same idea without requiring zero. The remainder on dividing $f(x)$ by $(x - a)$ is simply $f(a)$, so you can find a remainder by substitution rather than long division.

This is also useful in reverse: if you are told the remainder, set $f(a)$ equal to that remainder to form an equation for an unknown coefficient.

Factorising and solving a cubic

Once you have one factor, divide to reduce the cubic to a quadratic, which you can factorise by familiar methods.

Dividing polynomials

You can carry out the division by algebraic long division or by comparing coefficients. In the comparing-coefficients method, write the cubic as $(x - a)(x^2 + px + q)$, expand, and match the coefficients of each power of $x$ to find $p$ and $q$. Both methods are accepted; comparing coefficients is often quicker once you are confident, while long division is more mechanical and less error-prone under pressure.

To see comparing coefficients in action, suppose $x^3 + 2x^2 - 5x - 6$ has the factor $(x - 2)$, so $x^3 + 2x^2 - 5x - 6 = (x - 2)(x^2 + px + q)$. Expanding the right-hand side gives $x^3 + px^2 + qx - 2x^2 - 2px - 2q = x^3 + (p - 2)x^2 + (q - 2p)x - 2q$. Matching the $x^2$ coefficient, $p - 2 = 2$, so $p = 4$. Matching the constant, $-2q = -6$, so $q = 3$. The quadratic factor is therefore $x^2 + 4x + 3$, and checking the $x$ coefficient confirms it: $q - 2p = 3 - 8 = -5$, as required. Matching every coefficient, not just one, is the safeguard against an arithmetic slip.

Why this matters

Polynomials let you analyse cubic graphs, which have up to three roots and a characteristic S-shape with a local maximum and minimum that you can find by differentiation. The factor and remainder theorems also reappear whenever an equation will not factorise by inspection. Together with quadratic theory, this topic completes your toolkit for solving the algebraic equations that thread through the rest of the Pure unit.