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Module 2 · L4 of 15 ~35 min ⚡ +95 XP available

The Factor Theorem

The Remainder Theorem told you how to find the remainder — the Factor Theorem answers a deeper question: when is the remainder zero? If $P(a) = 0$, then $(x - a)$ is a factor. This single insight unlocks a systematic method for factorising any polynomial.

Today's hook — If $P(2) = 0$, the Remainder Theorem says the remainder when dividing $P(x)$ by $(x - 2)$ is zero. And if the remainder is zero, the division is exact — meaning $(x - 2)$ divides $P(x)$ perfectly, with no leftovers. That makes $(x - 2)$ a factor. By the end of this lesson you'll use this to fully factorise any cubic.
0/5QUESTS
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Recall — your gut answer first
+5 XP warm-up

If $P(2) = 0$, what does this tell you about the remainder when $P(x)$ is divided by $(x - 2)$? Before reading further — what do you think this implies about whether $(x - 2)$ is a factor of $P(x)$?

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The key idea
+5 XP to read

The Factor Theorem is a direct consequence of the Remainder Theorem with one extra condition. Zero remainder means exact division — which means $(x - a)$ is a factor.

Every Factor Theorem question asks you to do one of two things: test whether a given linear expression is a factor by checking if $P(a) = 0$, or fully factorise a polynomial by finding a factor first and then dividing to reduce the degree.

TEST A FACTOR P(a) = 0? FACTORISE FULLY divide + factor
$(x - a)$ is a factor $\Leftrightarrow P(a) = 0$
The factor, not the value
If $P(3) = 0$, then $(x - 3)$ is the factor — not $3$. The factor is the linear expression.
Try factors of the constant
For a monic polynomial $P(x)$, if $(x - a)$ is a factor then $a$ must divide the constant term. Try $\pm 1, \pm 2, \ldots$ first.
Divide then re-factorise
Once you find one factor, divide to get a lower-degree quotient, then factorise the quotient — usually by inspection or the quadratic formula.
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What you'll master
Know

Key facts

  • $(x - a)$ is a factor of $P(x)$ if and only if $P(a) = 0$
  • The Factor Theorem is the Remainder Theorem with $R = 0$
  • Integer roots of a monic polynomial divide the constant term
Understand

Concepts

  • Both directions of the proof ($\Rightarrow$ and $\Leftarrow$)
  • Why $a$ is called a zero and $(x - a)$ is called a factor
  • How the Factor Theorem connects polynomial roots to linear factors
Can do

Skills

  • Use the Factor Theorem to test whether a linear expression is a factor
  • Find a factor of a cubic by systematic substitution
  • Fully factorise a cubic using the Factor Theorem plus long division
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Key terms
Factor Theorem$(x - a)$ is a factor of $P(x)$ if and only if $P(a) = 0$.
Zero of a polynomialA value $a$ such that $P(a) = 0$; also called a root.
FactoriseWrite $P(x)$ as a product of simpler polynomials, e.g. linear and quadratic factors.
Quotient $Q(x)$The result after dividing $P(x)$ by $(x - a)$; has degree one less than $P(x)$.
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The Factor Theorem
core concept
$$\textbf{Factor Theorem: }(x - a) \text{ is a factor of } P(x) \Leftrightarrow P(a) = 0$$

Proof (both directions):

$(\Rightarrow)$ If $(x - a)$ is a factor: Then $P(x) = (x - a) \cdot Q(x)$ for some polynomial $Q(x)$. Substituting $x = a$:

$$P(a) = (a - a) \cdot Q(a) = 0 \cdot Q(a) = 0$$

$(\Leftarrow)$ If $P(a) = 0$: By the Remainder Theorem, the remainder when $P(x)$ is divided by $(x - a)$ is $P(a) = 0$. So:

$$P(x) = (x - a) \cdot Q(x) + 0 = (x - a) \cdot Q(x)$$

Therefore $(x - a)$ is a factor. ∎

Connection to the Remainder Theorem. The Factor Theorem is a special case: it's the Remainder Theorem applied when the remainder is exactly zero. This is why the two theorems are always taught together — each deepens the other.

Factor Theorem: (x - a) is a factor of P(x) P(a) = 0; Proof: substitute x = a into P(x) = (x-a)Q(x); Remainder Theorem with R = 0

Pause — copy the Factor Theorem into your book: $(x-a)$ is a factor of $P(x) \Leftrightarrow P(a) = 0$; it is the Remainder Theorem with remainder $R = 0$.

Quick check: $P(x) = x^3 - 3x^2 + 4$. We find $P(2) = 8 - 12 + 4 = 0$. Which statement follows directly from the Factor Theorem?

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Finding factors systematically
core concept

We just saw the Factor Theorem: $(x-a)$ is a factor of $P(x)$ if and only if $P(a) = 0$. That raises a question: to factorise a cubic, you need to find at least one root $a$ — what values of $a$ are worth trying first, and why? This card answers it → for monic integer polynomials, rational roots must be integer factors of the constant term; try $\pm 1$ and $\pm$ (factors of the constant) systematically.

To factorise a cubic $P(x) = x^3 + bx^2 + cx + d$, we need to find a value $a$ such that $P(a) = 0$. Since $a$ must divide the constant term $d$ (for monic integer polynomials), we try:

$$a \in \{\pm 1, \pm 2, \pm 3, \ldots\} \text{ — factors of the constant term } d$$

Once a zero $a$ is found:

  1. State: "Since $P(a) = 0$, by the Factor Theorem, $(x - a)$ is a factor."
  2. Divide $P(x)$ by $(x - a)$ to get quotient $Q(x)$.
  3. Factorise $Q(x)$ (usually a quadratic) by inspection or the quadratic formula.
  4. Write the final factorised form: $P(x) = (x - a) \cdot Q(x)$.
Why factors of the constant term? If $(x - a)$ is a factor of $x^n + \ldots + d$, then $P(a) = 0$. The constant term equals $(-a)^n \cdot (- \text{something}) + \ldots + d$ — for integer polynomials, $a$ must divide $d$. This limits your search to a small list.

To find a factor of a monic cubic: try (factors of constant term); Check P(a) = 0 — if yes, (x - a) is a factor by the Factor Theorem

Pause — copy the systematic root-finding method into your book: for a monic integer cubic, try $\pm 1$, $\pm d$ (factors of constant term $d$); when $P(a) = 0$, $(x-a)$ is a factor — divide out and factorise the quotient quadratic.

Did you get this? True or false: to find integer zeros of $P(x) = x^3 - 5x^2 + 2x + 8$, you should only test $\pm 1, \pm 2, \pm 4, \pm 8$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · TEST A FACTOR

Show that $(x - 2)$ is a factor of $P(x) = x^3 - 3x^2 + 4$.

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Evaluate $P(2)$: $P(2) = (2)^3 - 3(2)^2 + 4 = 8 - 12 + 4 = 0$
Substitute the zero of the divisor $(x - 2) \Rightarrow x = 2$.
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PROBLEM 2 · FACTORISE A CUBIC

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

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Constant term is $6$. Test factors: $\pm 1, \pm 2, \pm 3, \pm 6$.
For a monic polynomial, integer zeros divide the constant term.
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PROBLEM 3 · UNKNOWN COEFFICIENT

Find the value of $k$ if $(x - 2)$ is a factor of $P(x) = x^3 + kx^2 - 4x + 4$.

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If $(x - 2)$ is a factor, then by the Factor Theorem: $P(2) = 0$.
The condition for a factor is that the polynomial evaluates to zero at the zero of that factor.
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Fill the gap: To factorise $P(x) = x^3 - 6x^2 + 11x - 6$, we first test $P(1) =$ , which confirms $(x - 1)$ is a factor.

Misconceptions to fix · the 3 traps that cost marks

Trap 01
The zero vs the factor
If $P(3) = 0$, then $3$ is a zero of $P(x)$ — but the factor is $(x - 3)$, not $3$. Writing "3 is a factor" is wrong. A factor is always a polynomial (or number), and here it's the linear polynomial $(x - 3)$.
Trap 02
Not testing enough values
If $P(1) \ne 0$, don't give up. Try $P(-1)$, $P(2)$, $P(-2)$, etc. — all factors of the constant term. For $P(x) = x^3 - 7x + 6$, the constant term is 6, so try $\pm 1, \pm 2, \pm 3, \pm 6$.
Trap 03
Stopping at one factor
"Fully factorise" means expressing $P(x)$ as a product of factors that cannot be factorised further. After dividing by $(x - a)$, always factorise the quadratic quotient too — don't leave it as $x^2 + \ldots$

Did you get this? True or false: if $P(x) = (x - 1)(x^2 - 5x + 6)$, then the fully factorised form is $P(x) = (x - 1)(x - 2)(x - 3)$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

1

Use the Factor Theorem to show that $(x + 1)$ is a factor of $x^3 + 2x^2 - x - 2$. State your conclusion clearly.

2

Fully factorise $x^3 - 6x^2 + 11x - 6$. (Hint: try $x = 1, 2, 3$.)

3

Find the value of $k$ if $(x - 2)$ is a factor of $P(x) = x^3 + kx^2 - 4x + 4$.

4

Show that $(x + 2)$ is NOT a factor of $P(x) = x^3 - x^2 + x - 1$ by evaluating $P(-2)$.

5

Explain in your own words why the Factor Theorem only tells us about linear factors of the form $(x - a)$.

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Revisit your thinking

At the start, you were asked: if $P(2) = 0$, what does this tell you about $(x - 2)$ as a factor?

By the Remainder Theorem, $P(2) = 0$ means the remainder when $P(x)$ is divided by $(x - 2)$ is zero. Zero remainder means the division is exact — so $P(x) = (x - 2) \cdot Q(x)$ with no remainder. That is exactly the definition of a factor.

The Factor Theorem only tells us about linear factors because it relies on substituting a single value $a$ to make the factor zero. For a quadratic factor like $(x^2 - 5)$, you'd need to set $x^2 = 5$, giving $x = \pm\sqrt{5}$ — two values, not one — and the method is different.

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Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

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Short answer
ApplyBand 42 marks

Q1. Use the Factor Theorem to show that $(x + 1)$ is a factor of $x^3 + 2x^2 - x - 2$. (2 marks)

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ApplyBand 53 marks

Q2. Fully factorise $x^3 - 6x^2 + 11x - 6$. (3 marks)

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AnalyseBand 52 marks

Q3. Find the value of $k$ if $(x - 2)$ is a factor of $P(x) = x^3 + kx^2 - 4x + 4$. (2 marks)

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Comprehensive answers (click to reveal)

Activities: 1. $P(-1) = -1 + 2 + 1 - 2 = 0$; $(x+1)$ is a factor by FT · 2. $P(1) = 1-6+11-6=0$; $(x-1)$ is a factor; $Q = x^2-5x+6 = (x-2)(x-3)$; fully: $(x-1)(x-2)(x-3)$ · 3. $P(2)=0$: $8+4k-8+4=0 \Rightarrow 4k+4=0 \Rightarrow k=-1$ · 4. $P(-2)=-8-4-2-1=-15 \ne 0$; $(x+2)$ is not a factor · 5. The FT uses the Remainder Theorem; substituting one value makes a linear factor zero; a quadratic factor needs two values.

Q1 (2 marks): $P(-1) = (-1)^3 + 2(-1)^2 - (-1) - 2 = -1 + 2 + 1 - 2 = 0$ [1]. Since $P(-1) = 0$, by the Factor Theorem $(x + 1)$ is a factor [1].

Q2 (3 marks): $P(1) = 1 - 6 + 11 - 6 = 0$ ✓; $(x-1)$ is a factor [1]. $Q(x) = x^2 - 5x + 6 = (x-2)(x-3)$ [1]. $P(x) = (x-1)(x-2)(x-3)$ [1].

Q3 (2 marks): $P(2) = 0$: $8 + 4k - 8 + 4 = 0 \Rightarrow 4k = -4 \Rightarrow k = -1$ [2].

Odd one out: Three of these are correct conclusions from the Factor Theorem. Which one is wrong?

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Boss battle · The Factorisator
earn bronze · silver · gold

Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena
02
Science Jump · platform challenge

Climb platforms by answering Factor Theorem questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

Module overview · Extension 1 · Checkpoint 1 · Module quiz

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