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.
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.
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.
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:
This means the list of all possible rational zeros is:
How to generate the list:
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.
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?
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:
Polynomial division reminder: $P(x) = (x - a) \cdot Q(x) + R$. If $P(a) = 0$ then $R = 0$ and $P(x) = (x - a) \cdot Q(x)$.
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
Find all zeros of $P(x) = 2x^3 - 3x^2 - 11x + 6$.
Find all zeros of $P(x) = x^3 - 4x^2 + x + 6$.
Find all zeros of $P(x) = 2x^3 + x^2 - 5x + 2$.
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
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.
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?
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: $x = 2, -\tfrac{1}{2}, 1$.
Odd one out: Three of these values are zeros of $P(x) = x^3 - 6x^2 + 11x - 6$. Which one is NOT a zero?
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. List all possible rational zeros of $P(x) = 3x^3 + 2x^2 - 7x + 2$. (2 marks)
Q2. Find all zeros of $P(x) = x^3 - 4x^2 + x + 6$. Show your testing and division. (3 marks)
Q3. Find all zeros of $P(x) = 2x^3 + x^2 - 5x + 2$. Show all working including the polynomial division. (3 marks)
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].
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
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