NSWMaths Extension 1Syllabus dot point

How do we divide one polynomial by another and use the remainder and factor theorems to find roots?

Apply the division algorithm for polynomials and use the remainder and factor theorems to identify and verify factors and roots

A focused answer to the HSC Maths Extension 1 dot point on polynomial division. The division algorithm, the remainder theorem, the factor theorem, and using these to factorise cubics and quartics, with worked examples.

Generated by Claude OpusReviewed by Better Tuition Academy9 min answerUpdated 2026-05-20

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What this dot point is asking

NESA wants you to divide one polynomial by another (linear divisor at minimum, occasionally quadratic), state the result using the division algorithm, and use the remainder and factor theorems to identify roots without doing the full division when possible.

The answer

The division algorithm for polynomials

If $P(x)$ is a polynomial of degree $n$ and $D(x)$ is a polynomial of degree $k \le n$ , then there exist unique polynomials $Q(x)$ (the quotient) of degree $n - k$ and $R(x)$ (the remainder) of degree less than $k$ such that

P(x) = D(x) \, Q(x) + R(x).

When $D(x)$ is linear (degree $1$ ), the remainder $R(x)$ is a constant.

The remainder theorem

If $P(x)$ is divided by $(x - a)$ , the remainder is $P(a)$ . In other words,

P(x) = (x - a) Q(x) + P(a).

This lets you find the remainder without performing the division: just evaluate $P$ at $x = a$ .

For a divisor of the form $(b x - c)$ , the remainder is $P\!\left(\frac{c}{b}\right)$ .

The factor theorem

The factor theorem is the special case of the remainder theorem where the remainder is zero.

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

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

Long division of polynomials

To divide $P(x)$ by $D(x)$ , set up the long-division layout and at each step divide the leading term of the current dividend by the leading term of the divisor, multiply back, and subtract.

The integer-division analogy is exact. Stop when the remainder has lower degree than the divisor.

Strategy for factorising a cubic

To fully factorise a cubic $P(x)$ over the reals:

Find one rational root by trial. By the rational root theorem, any rational root $\frac{p}{q}$ (in lowest terms) has $p \mid$ constant term and $q \mid$ leading coefficient.
Use the factor theorem to confirm the root.
Divide $P(x)$ by the corresponding linear factor to get a quadratic quotient.
Factorise the quadratic by the standard methods.

If the quadratic factor has a negative discriminant, the cubic has one real root and two complex conjugate roots.

Worked example

Apply the remainder theorem

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

By the remainder theorem, the remainder is $P(-2) = 16 - 12 + 5 = 9$ .

Apply the factor theorem

Show that $(x - 3)$ is a factor of $P(x) = x^3 - 2 x^2 - 5 x + 6$ .

$P(3) = 27 - 18 - 15 + 6 = 0$ , so $(x - 3)$ is a factor.

Full factorisation

Factorise $P(x) = x^3 - 6 x^2 + 11 x - 6$ .

Test $x = 1$ : $P(1) = 1 - 6 + 11 - 6 = 0$ . So $(x - 1)$ is a factor.

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

Polynomial long division

Divide $P(x) = 2 x^3 + x - 4$ by $D(x) = x^2 - 1$ .

Layout: $2 x^3 + 0 x^2 + x - 4$ divided by $x^2 - 1$ .

First term: $\frac{2 x^3}{x^2} = 2 x$ . Multiply: $2 x (x^2 - 1) = 2 x^3 - 2 x$ . Subtract: $0 + 0 x^2 + 3 x - 4$ .

Second term: $\frac{0 x^2}{x^2} = 0$ . So the quotient is $2 x$ and the remainder is $3 x - 4$ .

Result: $2 x^3 + x - 4 = (x^2 - 1)(2 x) + (3 x - 4)$ .

Common traps

Sign error on the remainder theorem: IMATH_0 has remainder $P(a)$ . $P(x) \div (x + a)$ has remainder $P(-a)$ . Move the sign with care.
Forgetting placeholder zeros: When long-dividing, you must include zero coefficients for missing powers. $x^3 + 1$ is really $x^3 + 0 x^2 + 0 x + 1$ .
Using the wrong root from a $(b x - c)$ factor: IMATH_7 is a factor iff $P(1/2) = 0$ , not $P(1)$ or $P(2)$ .
Not testing the rational root candidates systematically: For $P(x) = 2 x^3 + 3 x^2 - 8 x + 3$ , candidates are $\pm 1, \pm 3, \pm \frac{1}{2}, \pm \frac{3}{2}$ . Try in order and confirm with the factor theorem.
Stopping at the linear divisor: A cubic factorises as a product of three linear factors over $\mathbb{R}$ only if all roots are real. If the quadratic quotient has no real roots, leave it as a quadratic factor.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2022 HSC Q32 marksWhen the polynomial

P(x) = x^3 - 2 x^2 + a x + 5

is divided by

(x - 2)

, the remainder is

7

. Find the value of

a

Show worked answer →

By the remainder theorem, $P(2) = 7$ .

$P(2) = 8 - 8 + 2 a + 5 = 2 a + 5 = 7$ , so $2 a = 2$ and $a = 1$ .

Markers reward the explicit statement of the remainder theorem, substitution of $x = 2$ , and a clean linear solve.

2020 HSC Q123 marksThe polynomial

P(x) = 2 x^3 + x^2 - 5 x + 2

has a factor of

(2 x - 1)

. Use this fact to fully factorise

P(x)

over the reals.

Show worked answer →

$(2 x - 1)$ is a factor, confirmed because $P\left(\frac{1}{2}\right) = 2 \cdot \frac{1}{8} + \frac{1}{4} - \frac{5}{2} + 2 = \frac{1}{4} + \frac{1}{4} - \frac{5}{2} + 2 = 0$ .

Divide $P(x)$ by $(2 x - 1)$ . Long division or inspection gives the quadratic quotient $x^2 + x - 2$ .

Factor the quadratic: $x^2 + x - 2 = (x + 2)(x - 1)$ .

$P(x) = (2 x - 1)(x + 2)(x - 1)$ .

Markers reward verifying the factor with the factor theorem, performing the division accurately, and producing a fully factored final form.