Apply the factor theorem and the remainder theorem to factorise polynomials and to solve polynomial equations

What this dot point is asking

VCAA wants you to use the factor and remainder theorems to factorise polynomials of degree $3$ or $4$ and to solve the resulting polynomial equations.

The remainder theorem

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

This lets you compute the remainder of a division by evaluating the polynomial at a point, without performing the division.

The factor theorem

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

Direct consequence of the remainder theorem (zero remainder means $(x - a)$ divides $P$ evenly).

Finding rational roots

For a polynomial with integer coefficients, the rational roots theorem says that any rational root $p/q$ (in lowest terms) has $p$ dividing the constant term and $q$ dividing the leading coefficient.

For $2x^3 + x^2 - 13x + 6$, candidates for rational roots are $\pm$ (divisors of $6$) over $\pm$ (divisors of $2$): $\pm 1, \pm 2, \pm 3, \pm 6, \pm 1/2, \pm 3/2$.

Test each by computing $P(a)$. Any value that gives zero is a root.

Polynomial long division

To divide $P(x)$ by $(x - a)$:

1. Divide the leading term of $P$ by $x$ to get the first term of the quotient.
2. Multiply $(x - a)$ by that term, subtract from $P$.
3. Repeat with the resulting remainder until the degree is less than $1$.

The result is $P(x) = (x - a) Q(x) + R$, where $R$ should be zero if $a$ is a root.

Synthetic division is an algorithmic shortcut for the same procedure with linear divisors $(x - a)$.

Standard procedure for factorising a cubic

1. Identify rational root candidates using the rational roots theorem.
2. Test by computing $P(a)$ at each candidate.
3. Once a root $a$ is found, divide $P$ by $(x - a)$ to get a quadratic.
4. Factor the quadratic (factorise, complete the square, or use the quadratic formula).
5. Combine to get the full factorisation.

Worked example (quartic)

$P(x) = x^4 - 5x^3 + 5x^2 + 5x - 6$.

Try $x = 1$: $1 - 5 + 5 + 5 - 6 = 0$. So $(x - 1)$ is a factor.

Divide: $P(x) = (x - 1)(x^3 - 4x^2 + x + 6)$.

For the cubic, try $x = -1$: $-1 - 4 - 1 + 6 = 0$. So $(x + 1)$ is a factor.

Divide: $x^3 - 4x^2 + x + 6 = (x + 1)(x^2 - 5x + 6)$.

Factor the quadratic: $x^2 - 5x + 6 = (x - 2)(x - 3)$.

Final: $P(x) = (x - 1)(x + 1)(x - 2)(x - 3)$.

Common traps

Sign error when stating the factor. Root $a = 2$ corresponds to factor $(x - 2)$, not $(x + 2)$.

Missing the rational-roots constraint. Trying $x = 5$ for $2x^3 + x^2 - 13x + 6$ wastes time; $5$ is not a divisor of $6$.

Wrong division procedure. Always check by multiplying the quotient by the divisor and confirming the original is recovered.

Treating a complex remainder as zero. Only zero remainder means $(x - a)$ is a factor.

In one sentence

The remainder theorem says $P(x)$ divided by $(x - a)$ leaves remainder $P(a)$; the factor theorem says $(x - a)$ is a factor of $P(x)$ if and only if $P(a) = 0$; combined with the rational roots theorem (rational root candidates are divisors of the constant term over divisors of the leading coefficient), these factor cubics and quartics into linear and quadratic factors.