Mathematical Methods · Units 3 & 4

Solving Polynomial Equations

Learn to solve polynomial equations the easy way with plain English intuition, the factor theorem, factorising cubics and quartics, the null factor law, worked examples and an auto marked practice test. VCE Maths Methods Units 3 and 4.

Hunting for the answer to a polynomial equation is detective work. You are handed a messy expression set equal to zero, and somewhere hidden inside are a few special numbers that make the whole thing collapse to nothing. Find those numbers and you have solved it. The brilliant trick of this topic is that you never have to guess wildly. There is a clean method that turns the hunt into a short, reliable recipe.

Why zero is the magic number

A polynomial equation is just a sum of powers of $x$ set equal to something. The first move is always the same. Get everything onto one side so the equation reads “stuff $= 0$ ”. Zero is special because of one beautifully simple fact: if you multiply numbers together and the result is zero, then at least one of those numbers must have been zero itself.

This is the null factor law, and it is the engine behind every solution in this lesson. Written out, it says that if $a \times b = 0$ , then either $a = 0$ or $b = 0$ . So once a polynomial is broken into a product of factors, each factor that equals zero hands you one solution.

(x - r_1)(x - r_2)(x - r_3) = 0 \quad \Rightarrow \quad x = r_1, \ x = r_2, \ x = r_3

The factor theorem: a shortcut to the first factor

Quadratics are easy to factorise, but a cubic or a quartic does not split apart so obviously. We need a way in. The factor theorem is that way in, and it is wonderfully direct.

The idea: if substituting $x = a$ into a polynomial $P(x)$ gives an answer of zero, then $(x - a)$ is a factor of that polynomial. In symbols:

P(a) = 0 \quad \Leftrightarrow \quad (x - a) \text{ is a factor of } P(x)

To find that first $a$ , you test small numbers that divide the constant term. For $x^3 - 7x + 6$ , the constant is $6$ , so you try $x = 1, -1, 2, -2, 3, -3$ until one of them lands on zero. Watch the sign carefully. A factor written as $(x - 2)$ comes from testing $x = 2$ , while $(x + 2)$ comes from testing $x = -2$ . Flipping that sign is the single most common mistake students make, so slow down on it.

From one factor to all the solutions

Once the factor theorem hands you one factor, the rest unfolds in a tidy sequence.

Test small values until $P(a) = 0$ , giving the first factor $(x - a)$ .
Divide the polynomial by $(x - a)$ to get a simpler quotient. Use long division or equate coefficients, whichever you find cleaner.
Factorise that quotient. For a cubic the quotient is a quadratic, which usually factorises by inspection.
Apply the null factor law to every factor and read off each solution.

A real exam favourite is the quartic with no odd powers, such as $x^4 - 5x^2 + 4$ . Treat $x^2$ as a single block, factorise it like a quadratic, then break each piece down again. Just remember that a factor like $x^2 + 4$ gives no real solution, because no real number squares to a negative.

Learn it, then fade the help

1 · I do — fully worked

Solve a cubic Solve $x^3 - 7x + 6 = 0$ . Test $x = 1$ : $1 - 7 + 6 = 0$ , so $(x - 1)$ is a factor. Divide: $x^3 - 7x + 6 = (x - 1)(x^2 + x - 6) = (x - 1)(x + 3)(x - 2)$ . Null factor law gives $x = 1, -3, 2$ .

2 · We do — your turn, guided

Same steps, your turn. Solve $x^3 - 2x^2 - 5x + 6 = 0$ . Test $x = 1$ : $1 - 2 - 5 + 6 = \;?$ — if it is $0$ then $(x - 1)$ is a factor; divide to find the quadratic.

Check your step

$x = 1$ gives $0$ , so $(x - 1)$ is a factor. Dividing: $(x - 1)(x^2 - x - 6) = (x - 1)(x - 3)(x + 2)$ , so $x = 1, 3, -2$ .

3 · You do — solo

No scaffolding now: solve $x^3 - 6x^2 + 11x - 6 = 0$ for all real $x$ .

Reveal solution

$x = 1$ gives $1 - 6 + 11 - 6 = 0$ , so $(x - 1)$ is a factor. Dividing: $(x - 1)(x^2 - 5x + 6) = (x - 1)(x - 2)(x - 3)$ , so $x = 1, 2, 3$ .

The cubic y = x³ - 7x + 6 crosses the x-axis exactly at its three solutions x = -3, 1 and 2, the same roots the factor theorem and null factor law produce.

The traps that cost marks

Two slips appear again and again in examiners’ reports. The first is rejecting a perfectly good solution, or keeping one that should be thrown out because of a hidden restriction such as a logarithm needing a positive input. Always sanity check each answer against the original problem. The second is stopping too soon. A cubic can have up to three real solutions and a quartic up to four, so if you have found only one, keep going. Find every factor, then claim every solution.

Lock it in with active recall

Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.

Active recall

Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).

State the null factor law.

State the factor theorem.

How is the remainder theorem related to the factor theorem?

How do you find the first factor of a cubic?

Why can a quartic have fewer than four real solutions?

Recall · Simultaneous Equations

For a line meeting a parabola, what does the discriminant decide?

Recall · Index and Logarithm Laws

How do you turn

e^{2x} - 5e^x + 6 = 0

into a quadratic?

See the recipe in action in the Worked Examples tab, then test yourself in Try It.

Worked Example 1Two functions meeting, from a real exam

Let $f(x) = x(x-2)(x+1)$ and let $g: R \to R$ , $g(x) = x - 2$ . Find the values of $x$ for which $f(x) = g(x)$ .

1
Set $f(x) = g(x)$ and bring everything to one side. Both sides share a common factor $(x-2)$ .
$x(x-2)(x+1) = x - 2 \implies (x-2)\big[x(x+1) - 1\big] = 0$
2
Expand the bracket fully.
$x(x+1) - 1 = x^2 + x - 1, \quad \text{so} \quad (x-2)(x^2 + x - 1) = 0$
3
Apply the null factor law to each factor.
$x - 2 = 0 \quad \text{or} \quad x^2 + x - 1 = 0$
4
Solve each. The quadratic needs the formula and gives an exact surd pair.
$x = 2, \quad x = \frac{-1 \pm \sqrt5}{2}$

Answer

x = \dfrac{-1 - \sqrt5}{2}, \ \dfrac{-1 + \sqrt5}{2}, \ 2

VCAA 2023 Mathematical Methods Exam 2, Section B Q1ci

Worked Example 2Verify a root, then factorise fully, from a real exam

Let $f: R \to R$ , $f(x) = x^3 - x^2 - 16x - 20$ . Verify that $x = 5$ is a solution of $f(x) = 0$ , then express $f(x)$ in the form $(x + d)^2(x - 5)$ , where $d \in R$ .

1
Substitute $x = 5$ into $f(x)$ to verify it is a root.
$f(5) = 125 - 25 - 80 - 20 = 0$
2
Since $f(5) = 0$ , the factor theorem says $(x - 5)$ is a factor. Divide $f(x)$ by $(x - 5)$ to find the quadratic factor.
$f(x) = (x - 5)(x^2 + 4x + 4)$
3
The quadratic is a perfect square, so factorise it.
$x^2 + 4x + 4 = (x + 2)^2$
4
Write $f(x)$ in the required form, so $d = 2$ . Working must be shown since this is worth more than one mark.
$f(x) = (x + 2)^2(x - 5)$

Answer

f(5) = 0, \ \text{ and } \ f(x) = (x+2)^2(x-5)

VCAA 2025 Mathematical Methods Exam 1, Q7a and Q7b

Worked Example 3Find a factor, then finish the job

Solve $x^3 - 7x + 6 = 0$ for all real values of $x$ .

1
Try small whole numbers that divide the constant $6$ . Test $x=1$ by substituting into the polynomial.
$(1)^3 - 7(1) + 6 = 1 - 7 + 6 = 0$
2
Since the result is $0$ , the factor theorem says $(x-1)$ is a factor. Divide to find the quadratic left over.
$x^3 - 7x + 6 = (x-1)(x^2 + x - 6)$
3
Factorise the quadratic by inspection.
$x^2 + x - 6 = (x+3)(x-2)$
4
Apply the null factor law: set each factor to zero.
$(x-1)(x+3)(x-2) = 0$

Answer

x = 1, \ x = -3, \ x = 2

Worked Example 4A cubic with a leading coefficient

Solve $2x^3 - 3x^2 - 11x + 6 = 0$ for all real values of $x$ .

1
Test factors of the constant first. Try $x=3$ .
$2(3)^3 - 3(3)^2 - 11(3) + 6 = 54 - 27 - 33 + 6 = 0$
2
So $(x-3)$ is a factor. Divide to find the quadratic factor.
$2x^3 - 3x^2 - 11x + 6 = (x-3)(2x^2 + 3x - 2)$
3
Factorise the quadratic. Here $2x^2 + 3x - 2 = (2x-1)(x+2)$ .
$(x-3)(2x-1)(x+2) = 0$
4
Set each factor to zero. The middle factor gives a fraction, so keep it.
$2x - 1 = 0 \implies x = \tfrac{1}{2}$

Answer

x = 3, \ x = \tfrac{1}{2}, \ x = -2

Worked Example 5A quartic that hides two quadratics

Solve $x^4 - 5x^2 + 4 = 0$ for all real values of $x$ .

1
There is no odd power, so treat $x^2$ as a single block. This is a quadratic in $x^2$ .
$(x^2)^2 - 5(x^2) + 4 = 0$
2
Factorise as if $x^2$ were the variable.
$(x^2 - 1)(x^2 - 4) = 0$
3
Factorise each difference of two squares fully.
$(x-1)(x+1)(x-2)(x+2) = 0$
4
Apply the null factor law to all four factors. A quartic can have up to four real solutions.
$x = 1, \ x = -1, \ x = 2, \ x = -2$

Answer

x = \pm 1, \ x = \pm 2

Practice test

Try it yourself

10 questions, 12 marks

Choose your answers, then submit to see your score and the full worked solutions. Multiple choice is marked for you, just like Exam 2 Section A.

Q1.Given that the graph of $y = g(x)$ , where $g(x) = \sqrt[3]{x - k} + m$ , passes through the origin, express $k$ in terms of $m$ .

1mark

Work this on paper. The worked solution appears once you submit.

VCAA 2024 Mathematical Methods Exam 1, Q8c

Q2.Find all values of $k$ such that the equation $x^2 + (4k+3)x + 4k^2 - \dfrac{9}{4} = 0$ has two real solutions for $x$ , one positive and one negative.

1mark

VCAA 2023 Mathematical Methods Exam 2, Section A Q19

Q3.Consider an algorithm that prints the roots of the cubic $f(x) = x^3 - 2x^2 - 9x + 18$ . It sets $c \leftarrow |f(0)|$ , then while $c > 0$ prints $c$ if $f(c) = 0$ and prints $-c$ if $f(-c) = 0$ , decrementing $c$ by $1$ each loop. In order, the algorithm prints the values:

1mark

VCAA 2024 Mathematical Methods Exam 2, Section A Q17

Q4.Consider the function $f: R \to R$ , $f(x) = (x + 1)(x + a)(x - 2)(x - 2a)$ where $a \in R$ . State, in terms of $a$ where required, the values of $x$ for which $f(x) = 0$ .

1mark

Work this on paper. The worked solution appears once you submit.

VCAA 2024 Mathematical Methods Exam 2, Section B Q1a

Q5.If $P(x) = x^3 + 2x^2 - 5x - 6$ , which of the following is a factor of $P(x)$ ?

1mark

Need a hint?

Use the factor theorem:

(x-a)

is a factor only when

P(a)=0

. Watch the sign,

(x+2)

means test

x=-2

Q6.The solutions to $(x+4)(2x-3) = 0$ are:

1mark

Need a hint?

Set each bracket to zero separately, then solve. Rearrange

2x-3=0

carefully to isolate

x

Q7.Given that $(x-1)$ is a factor of $x^3 - 6x^2 + 11x - 6$ , the full set of real solutions to $x^3 - 6x^2 + 11x - 6 = 0$ is:

1mark

Need a hint?

Divide out the given factor

(x-1)

first, then fully factorise the quadratic that is left before reading off every root.

Q8.How many distinct real solutions does $x^4 - 16 = 0$ have?

1mark

Need a hint?

Factorise as a difference of two squares, then ask which factors actually give real roots. A factor like

x^2+4

gives none.

Q9.When $P(x) = x^3 - 4x^2 + x + 6$ is divided by $(x+1)$ , the remainder is:

1mark

Need a hint?

Use the remainder theorem: the remainder on dividing by

(x+1)

P(-1)

, not

P(1)

Q10.Solve $x^3 - 2x^2 - 5x + 6 = 0$ for all real values of $x$ . Show every step of your working.

3marks

Work this on paper. The worked solution appears once you submit.

Frequently asked questions

How do I find the first factor of a cubic?

Use the factor theorem. Test small whole numbers that divide the constant term by substituting them into the polynomial. When one gives zero, the matching bracket is a factor. For example, if substituting two gives zero, then x minus two is a factor.

What is the difference between the factor theorem and the remainder theorem?

They are the same calculation read two ways. Substituting a value into the polynomial gives the remainder when you divide by the matching bracket. If that remainder happens to be zero, the bracket is a factor. So the factor theorem is just the special case of the remainder theorem where the remainder is zero.

Why does a quartic sometimes have fewer than four real solutions?

A quartic can have up to four real solutions, but some factors may have no real roots. A factor like x squared plus four can never be zero for a real number, because no real number squared gives a negative. Those factors are real and valid, they simply do not contribute real solutions.

Solving Polynomial Equations

Learn

Why zero is the magic number

The factor theorem: a shortcut to the first factor

From one factor to all the solutions

The traps that cost marks

Lock it in with active recall

Worked examples

Practice questions

Try it yourself

Frequently asked questions