The factor theorem and the remainder theorem for polynomial functions, the method of equating coefficients, and the factorisation of cubic and quartic polynomials over the rationals

What this dot point is asking

VCAA wants you to factorise polynomial expressions and solve polynomial equations using the factor and remainder theorems. The standard application is factorising a cubic with rational roots in Paper 1, which is non-negotiable algebra technique for the no-calculator paper.

The remainder theorem

When a polynomial $P(x)$ is divided by $(x - a)$, the remainder is $P(a)$.

This gives a quick way to find the remainder without doing the long division: just evaluate the polynomial at $x = a$.

Example. Find the remainder when $P(x) = x^3 + 2x^2 - 5x + 4$ is divided by $(x - 2)$.

$P(2) = 8 + 8 - 10 + 4 = 10$. The remainder is $10$.

The factor theorem

$(x - a)$ is a factor of $P(x)$ if and only if $P(a) = 0$.

This is the special case of the remainder theorem with remainder zero. It is the workhorse for cubic and quartic factorisation in Paper 1.

Rational root candidates. For a polynomial with integer coefficients, the rational root theorem says that any rational root $p/q$ in lowest terms must have $p$ dividing the constant term and $q$ dividing the leading coefficient. So for $P(x) = 2x^3 - 3x^2 + 4x - 6$, the candidates are $\pm 1, \pm 2, \pm 3, \pm 6, \pm 1/2, \pm 3/2$.

The standard cubic factorisation method

Factorise a cubic $P(x)$ in three steps.

Step 1: find a root by trial

Test small integers (especially $\pm 1$ and factors of the constant term). Substitute into $P$ and check for zero.

Step 2: divide out the factor

Once you have a root $a$, divide $P(x)$ by $(x - a)$ to get a quadratic quotient. Use polynomial long division or synthetic division.

Step 3: factorise the quadratic

Use any of the standard quadratic methods: factor, complete the square, or the quadratic formula. If the discriminant is negative, the quadratic is irreducible over the reals and you stop at $(x - a) (\text{irreducible quadratic})$.

Worked example

Factorise $P(x) = x^3 + x^2 - 8x - 12$ over the rationals.

Step 1. Trial. $P(-2) = -8 + 4 + 16 - 12 = 0$. So $(x + 2)$ is a factor.

Step 2. Divide:

Equating coefficients on the right: $x^3$ coefficient is $1$, OK. $x^2$ coefficient is $b + 2 = 1$, so $b = -1$. Constant is $2c = -12$, so $c = -6$. Check $x$ coefficient: $c + 2b = -6 - 2 = -8$. OK.

Step 3. $x^2 - x - 6 = (x - 3)(x + 2)$.

Final: $P(x) = (x + 2)^2 (x - 3)$. (Note the repeated root at $x = -2$.)

Equating coefficients

The method of equating coefficients is the algebraic alternative to long division. Write the expected factored form with unknown coefficients, multiply out, and match coefficients of each power of $x$ on both sides.

Example. If $P(x) = x^3 - 5x^2 + 4x + 12$ has $(x - 3)$ as a factor, write $P(x) = (x - 3)(x^2 + b x + c)$. Multiply out: $x^3 + b x^2 + c x - 3 x^2 - 3 b x - 3 c = x^3 + (b - 3) x^2 + (c - 3b) x - 3c$.

Match: $b - 3 = -5$ gives $b = -2$. $-3 c = 12$ gives $c = -4$. Check $c - 3b = -4 + 6 = 2$, but the original has coefficient $4$. So $(x - 3)$ is not actually a factor. Confirm: $P(3) = 27 - 45 + 12 + 12 = 6 \neq 0$.

Quartics

For a quartic $P(x)$, the same approach extends:

1. Find one rational root by trial, factor out $(x - a)$ to get a cubic.
2. Repeat for the cubic to get another linear factor.
3. Factorise the resulting quadratic.

If the quartic has the special form $a(x^2)^2 + b(x^2) + c$ (a "biquadratic"), substitute $u = x^2$ and factorise as a quadratic in $u$, then solve $x^2 = \text{root}$ for $x$.

Common Paper 1 traps

Forgetting the rational root theorem candidates. Only test rational numbers $p/q$ where $p$ divides the constant and $q$ divides the leading coefficient.

Sign errors in long division. Each subtraction step changes signs. Write out the division carefully.

Stopping at the cubic factorisation. "Fully factorise" means continue until each factor is linear or irreducible quadratic.

Confusing the remainder theorem direction. Dividing by $(x - a)$ uses $P(a)$, not $P(-a)$. Dividing by $(x + 3)$ uses $P(-3)$.

Missing repeated roots. When the quadratic quotient shares a root with the original linear factor, you get a repeated root. Always check.

In one sentence

The factor theorem says $(x - a)$ divides $P(x)$ exactly when $P(a) = 0$, the remainder theorem says the remainder on division by $(x - a)$ is $P(a)$, and together they give the standard Paper 1 method to factorise cubics and quartics by finding one rational root, dividing it out, and factorising the resulting lower-degree polynomial.