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Module 2 · L5 of 15 ~35 min ⚡ +95 XP available

Finding Zeros of Polynomials

A structural engineer must know precisely where a beam deflection equation equals zero — because that pinpoints the failure point. A chemist finds the concentration at which a reaction rate drops to zero. In Extension 1, zeros of polynomials are the foundation for factorisation, graphing, and solving higher-degree equations. In this lesson you'll master the Rational Root Theorem and use the Factor Theorem to systematically crack any cubic or quartic polynomial.

Today's hook — For $P(x) = 2x^3 - 5x^2 + x + 2$, there are infinitely many numbers you could test as zeros. How do you know which ones to try without guessing forever? The Rational Root Theorem gives you a finite list of candidates — and the Factor Theorem tells you the moment you've found one. By the end of this lesson you'll have a two-step system that finds all zeros without luck.

0/5QUESTS

Recall — your gut answer first

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If $P(x) = 2x^3 - 5x^2 + x + 2$, what are the possible rational zeros? Think about the factors of the constant term ($2$) divided by the factors of the leading coefficient ($2$). Write down all the candidates you can think of — don't look it up yet.

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The two moves

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Every zero-finding question in this module uses exactly two ideas. Lock list candidates with the Rational Root Theorem and test and factor using the Factor Theorem into memory and you'll never be stuck.

Every polynomial zero problem lives on one of two roads: generate a finite list of rational candidates ($\pm p/q$) using the Rational Root Theorem, then test each candidate and divide out factors until the polynomial is fully factorised.

$P(a) = 0 \Rightarrow (x - a) \mid P(x)$

Rational Root Theorem

If $p/q$ is a rational zero, then $p \mid a_0$ (constant term) and $q \mid a_n$ (leading coefficient). Always list ALL candidates before testing.

Factor Theorem shortcut

Test small integers first ($\pm 1, \pm 2, \ldots$) — they are the most likely zeros. Once $P(a)=0$, divide out $(x-a)$ and solve the reduced polynomial.

Not all zeros are rational

The Rational Root Theorem only identifies possible rational zeros. If none work, the zeros are irrational or complex — don't keep testing forever.

What you'll master

Know

Key facts

The Rational Root Theorem: if $p/q$ is a rational zero, then $p \mid a_0$ and $q \mid a_n$
The Factor Theorem: $P(a) = 0$ if and only if $(x - a)$ is a factor of $P(x)$
Why the list of rational candidates is always finite

Understand

Concepts

Why we only need to test rationals of the form $\pm p / q$
The connection between zeros, factors, and roots of the polynomial equation
Why many polynomials have no rational zeros at all

Can do

Skills

List all possible rational zeros of any polynomial with integer coefficients
Systematically find all rational zeros of cubic and quartic polynomials
Divide out each found factor and factorise the reduced polynomial

Key terms

Zero (root)A value $a$ where $P(a) = 0$. Also called a root of the equation $P(x) = 0$.

Rational Root TheoremIf $P(x)$ has integer coefficients and $p/q$ (in lowest terms) is a rational zero, then $p$ divides the constant term $a_0$ and $q$ divides the leading coefficient $a_n$.

Factor Theorem$(x - a)$ is a factor of $P(x)$ if and only if $P(a) = 0$. This is the test that confirms a zero.

Monic polynomialA polynomial with leading coefficient 1. For monic polynomials, $q = 1$ and the only rational zeros are integer factors of the constant term.

Polynomial divisionAfter finding a zero $a$, dividing $P(x)$ by $(x - a)$ gives a polynomial of degree one less, which is easier to factorise.

Complete factorisationWriting $P(x)$ as a product of linear (and irreducible quadratic) factors over the rationals, giving all zeros simultaneously.

The Rational Root Theorem

core concept

For any polynomial $P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$ with integer coefficients, the Rational Root Theorem states:

$$\text{If } \frac{p}{q} \text{ (in lowest terms) is a rational zero, then } p \mid a_0 \text{ and } q \mid a_n$$

This means the list of all possible rational zeros is:

$$\text{possible rational zeros} = \pm\frac{\text{factors of } a_0}{\text{factors of } a_n}$$

How to generate the list:

Write down all positive factors of $|a_0|$ (numerators).
Write down all positive factors of $|a_n|$ (denominators).
Form every combination $\pm p/q$, eliminating duplicates.

Example: For $P(x) = 2x^3 - 3x^2 - 11x + 6$: factors of $6$ are $1, 2, 3, 6$; factors of $2$ are $1, 2$. Possible zeros: $\pm 1, \pm 2, \pm 3, \pm 6, \pm \tfrac{1}{2}, \pm \tfrac{3}{2}$. That is 12 candidates — a finite, manageable list.

Why it works. If you substitute $x = p/q$ into $P(x) = 0$ and multiply through by $q^n$, then $p$ must divide the constant term (since all other terms are divisible by $p$) and $q$ must divide the leading coefficient (since all other terms are divisible by $q$). The proof uses divisibility properties of integers.

Rational Root Theorem: for P(x) with integer coefficients, rational zeros are p/q where p a_0 and q a_n; For monic polynomials (a_n = 1), rational zeros are just (integer factors of a_0)

Pause — copy the Rational Root Theorem into your book: rational zeros of $a_n x^n + \cdots + a_0$ are $p/q$ with $p \mid a_0$, $q \mid a_n$; for monic integer polynomials, rational zeros are integer factors of $a_0$.

Quick check: For $P(x) = 3x^3 + 5x^2 - 2x + 4$, which of the following is a possible rational zero according to the Rational Root Theorem?

Testing candidates and factorising

core concept

We just saw the Rational Root Theorem: rational zeros of a polynomial are of the form $p/q$ where $p$ divides $a_0$ and $q$ divides $a_n$. That raises a question: after listing candidates, how do you efficiently test them and then completely factorise once you find a zero? This card answers it → test $\pm 1$ first (simplest), then use the Factor Theorem; when $P(a) = 0$, divide by $(x-a)$ and factorise the quotient.

Once you have your list of candidates, you test each one using the Factor Theorem: if $P(a) = 0$, then $(x - a)$ is a factor.

The systematic process:

Test candidates in order: $\pm 1$ first (usually fastest), then $\pm 2$, then fractions.
When $P(a) = 0$, perform polynomial long division or synthetic division to divide $P(x)$ by $(x - a)$.
Factorise the resulting polynomial of reduced degree.
Repeat until the polynomial is fully factorised into linear (and possibly irreducible quadratic) factors.

Polynomial division reminder: $P(x) = (x - a) \cdot Q(x) + R$. If $P(a) = 0$ then $R = 0$ and $P(x) = (x - a) \cdot Q(x)$.

$$P(x) = (x - a_1)(x - a_2) \cdots (x - a_k) \cdot Q(x)$$

where $a_1, a_2, \ldots, a_k$ are the rational zeros found, and $Q(x)$ contains any remaining irrational or complex zeros.

Test 1 first — they are the simplest and most common zeros; When P(a) = 0: divide P(x) (x - a) using long division or inspection

Pause — copy the testing protocol into your book: test $\pm 1$ first — they are the simplest; when $P(a) = 0$: divide $P(x)$ by $(x-a)$ using long division or inspection to get the quotient for further factorisation.

Did you get this? True or false: if the Rational Root Theorem produces no zeros, the polynomial definitely has no real zeros at all.

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

PROBLEM 1 · CUBIC WITH INTEGER ZEROS

Find all zeros of $P(x) = 2x^3 - 3x^2 - 11x + 6$.

Possible rational zeros: $\pm \dfrac{\text{factors of }6}{\text{factors of }2} = \pm 1, \pm 2, \pm 3, \pm 6, \pm \dfrac{1}{2}, \pm \dfrac{3}{2}$

Apply the Rational Root Theorem: numerators divide the constant term 6, denominators divide the leading coefficient 2.

PROBLEM 2 · MONIC CUBIC

Find all zeros of $P(x) = x^3 - 4x^2 + x + 6$.

$P(x)$ is monic, so possible integer zeros: $\pm 1, \pm 2, \pm 3, \pm 6$

For monic polynomials ($a_n = 1$), the Rational Root Theorem simplifies — only integer factors of the constant term are possible rational zeros.

PROBLEM 3 · FRACTIONAL ZERO

Find all zeros of $P(x) = 2x^3 + x^2 - 5x + 2$.

Possible rational zeros: $\pm 1, \pm 2, \pm \dfrac{1}{2}$

Factors of constant term 2: $\{1, 2\}$. Factors of leading coefficient 2: $\{1, 2\}$. All $\pm p/q$ combinations.

Fill the gap: For $P(x) = x^3 - 6x^2 + 11x - 6$, since it is monic the possible rational zeros are $\pm$ (integer factors of the constant term). The constant term is so the candidates are $\pm 1, \pm 2, \pm 3, \pm 6$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Thinking every polynomial has rational zeros

The Rational Root Theorem gives possible rational zeros. If none of the candidates satisfy $P(a) = 0$, the polynomial has no rational zeros — it doesn't mean you've made an error. For example, $P(x) = x^2 - 2$ has zeros $\pm\sqrt{2}$, which are irrational. The list $\pm 1, \pm 2$ won't work, and that's correct.

Trap 02

Missing fractions in the candidate list

Students often list only integer candidates and forget fractions like $\pm \tfrac{1}{2}$ or $\pm \tfrac{3}{2}$. For a non-monic polynomial like $2x^3 + \ldots$, the denominators must include factors of the leading coefficient. Missing these means missing valid zeros — as in Worked Example 3 where $x = \tfrac{1}{2}$ was a zero.

Trap 03

Not completing the factorisation

Finding one zero and stopping is an incomplete answer. After dividing out $(x - a)$, you must factorise the quotient polynomial completely. In a cubic, finding one zero gives a quadratic quotient — which must then be factorised or solved using the quadratic formula to find all three zeros.

Did you get this? True or false: for $P(x) = 4x^3 - 3x + 1$, the value $x = \dfrac{1}{2}$ is a possible rational zero according to the Rational Root Theorem.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

List all possible rational zeros of $P(x) = 3x^3 + 2x^2 - 7x + 2$. How many candidates are there?

Find all zeros of $P(x) = x^3 - 4x^2 + x + 6$. Show all your testing.

Find all zeros of $P(x) = 2x^3 + x^2 - 5x + 2$. Show your full working including the division step.

For $P(x) = x^3 - 7x + 6$: (a) list candidates, (b) find all zeros, (c) write $P(x)$ in fully factorised form.

Explain why the Rational Root Theorem is useful even though it doesn't guarantee finding any zeros. What should you do if none of the candidates work?

Revisit your thinking

Earlier you tried to list possible rational zeros of $P(x) = 2x^3 - 5x^2 + x + 2$. Using the Rational Root Theorem: factors of $2$ are $\{1, 2\}$; factors of $2$ (leading coefficient) are $\{1, 2\}$. The candidates are $\pm 1, \pm 2, \pm \tfrac{1}{2}$.

Testing: $P(\tfrac{1}{2}) = 2(\tfrac{1}{8}) - 5(\tfrac{1}{4}) + \tfrac{1}{2} + 2 = \tfrac{1}{4} - \tfrac{5}{4} + \tfrac{2}{4} + \tfrac{8}{4} = \tfrac{6}{4} \ne 0$. Try $P(2) = 16 - 20 + 2 + 2 = 0$ ✓. So $(x - 2)$ is a factor, giving $P(x) = (x - 2)(2x^2 - x - 1) = (x - 2)(2x + 1)(x - 1)$. Zeros: $x = 2, -\tfrac{1}{2}, 1$.

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Odd one out: Three of these values are zeros of $P(x) = x^3 - 6x^2 + 11x - 6$. Which one is NOT a zero?

Multiple choice

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

Short answer

ApplyBand 42 marks

Q1. List all possible rational zeros of $P(x) = 3x^3 + 2x^2 - 7x + 2$. (2 marks)

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ApplyBand 43 marks

Q2. Find all zeros of $P(x) = x^3 - 4x^2 + x + 6$. Show your testing and division. (3 marks)

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ApplyBand 4–53 marks

Q3. Find all zeros of $P(x) = 2x^3 + x^2 - 5x + 2$. Show all working including the polynomial division. (3 marks)

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Comprehensive answers (click to reveal)

Activity 1: 1. Factors of 2: $\{1,2\}$; factors of 3: $\{1,3\}$; candidates: $\pm 1, \pm 2, \pm \tfrac{1}{3}, \pm \tfrac{2}{3}$ — 8 candidates. 2. Monic, so $\pm 1, \pm 2, \pm 3, \pm 6$. Test $P(1)=4 \ne 0$, $P(-1)=0$ ✓. Divide: $(x+1)(x^2-5x+6) = (x+1)(x-2)(x-3)$. Zeros: $x = -1, 2, 3$. 3. Candidates: $\pm 1, \pm 2, \pm \tfrac{1}{2}$. $P(1) = 0$. Divide: $(x-1)(2x^2+3x-2) = (x-1)(2x-1)(x+2)$. Zeros: $x = 1, \tfrac{1}{2}, -2$. 4. Candidates: $\pm 1, \pm 2, \pm 3, \pm 6$. $P(1)=0$. Divide: $(x-1)(x^2-6x+6)$. Wait — check: $(x-1)(x^2-6x+6)=x^3-6x^2+6x-x^2+6x-6=x^3-7x^2+12x-6 \ne P(x)$. Retry: $P(x)=x^3-7x+6$. $P(1)=1-7+6=0$ ✓. Divide: $(x-1)(x^2+x-6) = (x-1)(x-2)(x+3)$. Zeros: $x=1,2,-3$. 5. It is useful because it reduces infinite guessing to a finite list. If none work, the polynomial has no rational zeros and you would use the quadratic formula (after trying all candidates) or declare the zeros irrational/complex.

Q1 (2 marks): Factors of $2$: $\{1,2\}$ [0.5]; factors of $3$: $\{1,3\}$ [0.5]. Possible zeros: $\pm 1, \pm 2, \pm \tfrac{1}{3}, \pm \tfrac{2}{3}$ [1]. (8 candidates total.)

Q2 (3 marks): Monic: candidates $\pm 1, \pm 2, \pm 3, \pm 6$ [0.5]. $P(-1) = -1-4-1+6 = 0$ ✓ [0.5]. Divide: $P(x) = (x+1)(x^2-5x+6)$ [1]. Factorise: $(x+1)(x-2)(x-3)$; zeros $x = -1, 2, 3$ [1].

Q3 (3 marks): Candidates: $\pm 1, \pm 2, \pm \tfrac{1}{2}$ [0.5]. $P(1) = 2+1-5+2 = 0$ ✓ [0.5]. Divide: $P(x) = (x-1)(2x^2+3x-2)$ [1]. Factorise: $(x-1)(2x-1)(x+2)$; zeros $x = 1, \tfrac{1}{2}, -2$ [1].

Boss battle · The Zero Hunter

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

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering polynomial zero questions. Lighter alternative to the boss.

Mark lesson as complete

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