MathsICSE Class 10 · Factorisation of Polynomials

Factor Theorem: Statement, Unknown Coefficients and Solved Questions (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

The factor theorem says (x − a) is a factor of a polynomial f(x) exactly when f(a) = 0. For a factor (px − q), test f(q/p) = 0. To find an unknown coefficient, set f(a) = 0 and solve; with two unknowns and two factors, form two equations and solve them together.

Turn the factor into zero to get x = a, then substitute: f(a) = 0 means it is a factor, anything else means it is not.

Illustration drawn for QuestionBuddy.

Understand Factor Theorem

The factor theorem turns a question about division into a question about substitution.

Statement. For a polynomial $f (x)$ and a number $a$ :

$(x - a) is a factor of f (x) ⟺ f (a) = 0.$

It follows from the remainder theorem: dividing $f (x)$ by $(x - a)$ leaves remainder $f (a)$ , and a divisor is a factor exactly when the remainder is zero.

Which value to substitute. Set the factor equal to zero and solve:

$(x - 3)$ : substitute $x = 3$ .
$(x + 2)$ : substitute $x = - 2$ .
$(2 x - 1)$ : substitute $x = \frac{1}{2}$ .
$(3 x + 2)$ : substitute $x = - \frac{2}{3}$ .

Three kinds of question.

1. Test a factor. Work out $f (a)$ . If it is 0, the factor is confirmed; otherwise it is not a factor. Write the conclusion in words.
2. Find one unknown. If $(x - 1)$ is a factor of $3 x^{3} + 8 x^{2} - 15 x + k$ (2024 Q8(i)), then $f (1) = 3 + 8 - 15 + k = 0$ , so $k = 4$ . The examiners reported calculation slips giving $k = - 4$ .
3. Find two unknowns. If two factors are given, each gives one equation; solve the pair simultaneously. Check by substituting both values back.

If $x^{2} - a^{2}$ is a factor, then both $(x - a)$ and $(x + a)$ are factors, so you get two equations from one statement.

Finding a factor to start factorising. For a cubic with integer coefficients and leading coefficient 1, try divisors of the constant term ( $\pm 1, \pm 2, \pm 3, \dots$ ). For a leading coefficient other than 1, also try fractions $\frac{divisor of constant}{divisor of leading coefficient}$ . Once one factor is found, divide (or compare coefficients) to get a quadratic and factorise it; that is the topic on factorising a cubic completely.

Arithmetic care. Most lost marks here are slips with signs and powers: $(- 2)^{3} = - 8$ , $(- 2)^{2} = 4$ , $(\frac{1}{2})^{3} = \frac{1}{8}$ . Write each term of $f (a)$ separately before adding.

Factor Theorem: formulas to remember

Factor theorem: $(x - a) ∣ f (x) ⟺ f (a) = 0$; ' $∣$ ' means 'is a factor of'.
Linear factor with a coefficient: $(p x - q) ∣ f (x) ⟺ f (\frac{q}{p}) = 0$; Set the factor to zero to find the value.
Factor x + a: $(x + a) ∣ f (x) ⟺ f (- a) = 0$; Note the sign change.
Quadratic factor x² − a²: $(x^{2} - a^{2}) ∣ f (x) ⟺ f (a) = 0 and f (- a) = 0$; Two conditions from one statement.
Possible integer roots: $f (x) = x^{3} + \dots + c : try x = \pm divisors of c$; For leading coefficient 1.

How to solve Factor Theorem questions

Substitute carefully, then solve the equation(s) you get.

1Set the given factor equal to zero to find the value to substitute.
2Write $f (value)$ term by term, with powers of negative numbers and fractions in brackets.
3To test a factor: if $f (a) = 0$ , write 'hence $(x - a)$ is a factor'; otherwise 'it is not a factor'.
4To find one unknown: set $f (a) = 0$ and solve.
5To find two unknowns: write one equation for each factor and solve simultaneously.
6Check by substituting your values back.
7If asked, go on to factorise completely.

Factor Theorem in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Factor theorem to find break-even points

Suppose a small company models its profit (in lakh rupees) as $P (x) = x^{3} - 6 x^{2} + 11 x - 6$ , where $x$ is the number of thousand units sold, for $0 \leq x \leq 4$ .

Show that the company breaks even ( $P = 0$ ) at $x = 1$ , and find the other break-even points.

Try it first, then show the solutionSolution

$P (1) = 1 - 6 + 11 - 6 = 0$ , so $(x - 1)$ is a factor.
$P (x) = (x - 1) (x^{2} - 5 x + 6) = (x - 1) (x - 2) (x - 3)$ .
$P = 0$ at $x = 1, 2, 3$ .

Answer: Break-even at 1, 2 and 3 thousand units.

Memory hook: Zero of the polynomial = a factor of the polynomial.

Dimensions from a volume expression

Imagine a box designer whose box has volume $2 x^{3} + 7 x^{2} + 7 x + 2$ cm³, with each edge a linear expression in $x$ .

Find the three edges, and the volume when $x = 3$ .

Try it first, then show the solutionSolution

$f (- 1) = - 2 + 7 - 7 + 2 = 0$ , so $(x + 1)$ is a factor.
$f (x) = (x + 1) (2 x^{2} + 5 x + 2) = (x + 1) (2 x + 1) (x + 2)$ .
At $x = 3$ : edges 4, 7, 5 cm; volume $= 140$ cm³.

Answer: Edges $(x + 1)$ , $(2 x + 1)$ , $(x + 2)$ cm; 140 cm³ when $x = 3$ .

Memory hook: One zero unlocks the whole factorisation.

Tricks to remember Factor Theorem

Short memory aids. Say each one aloud once, then use it on a question the same day.

Factor to zero, then substitute
Whatever the factor, set it equal to 0 and solve: that is the value to put in.
Use it when: Avoiding sign errors with $(x + a)$ and $(p x - q)$ .
Zero remainder = factor
Factor theorem is the remainder theorem with remainder 0.
Use it when: Remembering why it works.
Two factors, two equations
Each given factor gives one equation; two unknowns need two equations.
Use it when: Finding $a$ and $b$ .

Factor Theorem: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Spotting a factor · 1 mark · MCQBoard level

Which of the following is a factor of $x^{3} - 2 x^{2} - 5 x + 6$ ?
(a) $x + 1$
(b) $x - 1$
(c) $x + 3$
(d) $x - 2$

Try it first, then show the solutionSolution

$f (1) = 1 - 2 - 5 + 6 = 0$ .
(Check others: $f (- 1) = 8$ , $f (- 3) = - 24$ , $f (2) = - 4$ .)

Answer: (b) $x - 1$

Example 2: Testing a factor with a coefficient · 3 marks · Short answerBoard level

Show that $(2 x - 1)$ is a factor of $2 x^{3} + x^{2} - 7 x + 3$ .

Try it first, then show the solutionSolution

$2 x - 1 = 0 \Rightarrow x = \frac{1}{2}$ .
$f (\frac{1}{2}) = 2 \cdot \frac{1}{8} + \frac{1}{4} - \frac{7}{2} + 3 = \frac{1}{4} + \frac{1}{4} - \frac{7}{2} + 3 = 0$ .
Since $f (\frac{1}{2}) = 0$ , $(2 x - 1)$ is a factor.

Answer: $f (\frac{1}{2}) = 0$ , so $(2 x - 1)$ is a factor.

Example 3: One unknown · 3 marks · Short answerHarder board level

Find $k$ if $(x + 2)$ is a factor of $x^{3} + k x^{2} - x - 6$ .

Try it first, then show the solutionSolution

$f (- 2) = (- 2)^{3} + k (- 2)^{2} - (- 2) - 6 = - 8 + 4 k + 2 - 6$ .
$4 k - 12 = 0 \Rightarrow k = 3$ .

Answer: $k = 3$

Example 4: Two unknowns · 4 marks · Short answerHarder board level

$(x - 1)$ and $(x + 3)$ are factors of $x^{3} + a x^{2} + b x - 6$ . Find $a$ and $b$ , and the third factor.

Try it first, then show the solutionSolution

$f (1) = 1 + a + b - 6 = 0 \Rightarrow a + b = 5$ .
$f (- 3) = - 27 + 9 a - 3 b - 6 = 0 \Rightarrow 3 a - b = 11$ .
Adding: $4 a = 16$ , $a = 4$ , $b = 1$ .
$x^{3} + 4 x^{2} + x - 6 = (x - 1) (x + 3) (x + 2)$ ; the third factor is $(x + 2)$ .

Answer: $a = 4$ , $b = 1$ ; third factor $(x + 2)$ .

Example 5: The unknown is also the root · 3 marks · Short answerHOTS / competency

Find $a$ if $(x - a)$ is a factor of $x^{3} - a x^{2} + 2 x + a - 1$ .

Try it first, then show the solutionSolution

$f (a) = a^{3} - a \cdot a^{2} + 2 a + a - 1 = a^{3} - a^{3} + 3 a - 1$ .
$3 a - 1 = 0 \Rightarrow a = \frac{1}{3}$ .

Answer: $a = \frac{1}{3}$

Example 6: Case study: a crate's dimensions · 4 marks · Case-basedHOTS / competency

The volume of a crate is $V (x) = x^{3} + 6 x^{2} + 11 x + 6$ cubic units, and each edge is a linear expression in $x$ .
(a) Show that $(x + 1)$ is a factor of $V (x)$ .
(b) Find all three edges.
(c) Find the edges and volume when $x = 4$ .

Try it first, then show the solutionSolution

(a) $V (- 1) = - 1 + 6 - 11 + 6 = 0$ , so $(x + 1)$ is a factor.
(b) $V (x) = (x + 1) (x^{2} + 5 x + 6) = (x + 1) (x + 2) (x + 3)$ .
(c) Edges 5, 6, 7; volume $= 210$ (check: $64 + 96 + 44 + 6 = 210$ ).

Answer: (a) $V (- 1) = 0$ (b) $(x + 1), (x + 2), (x + 3)$ (c) 5, 6, 7 units; 210 cubic units.

Where students lose marks in Factor Theorem

The slips examiners see most often, and what to write instead.

Mistake: Substituting the wrong sign
For $(x + 2)$ the value is $- 2$ , not $2$ .
Wrong working
$(x + 2)$ is a factor, so $f (2) = 0$ .
Right working
$f (- 2) = 0$ .
Do this: Set the factor equal to zero first.
Usually costs: Every mark of the question.
Mistake: Calculation slips with powers
Odd powers of negatives are negative; fractions must be cubed fully.
Wrong working
$(- 2)^{3} = 8$ ; $2 (\frac{1}{2})^{3} = 1$ .
Right working
$(- 2)^{3} = - 8$ ; $2 \cdot \frac{1}{8} = \frac{1}{4}$ .
Do this: Write powers in brackets and evaluate one term at a time.
Usually costs: Reported for 2024 Q8(i) and 2025 Q5(ii).
Mistake: No conclusion in words
The method mark needs the link 'f(a) = 0, hence a factor'.
Wrong working
$f (\frac{1}{2}) = 0$ . (Stops.)
Right working
$f (\frac{1}{2}) = 0$ , hence $(2 x - 1)$ is a factor.
Do this: Always finish with a sentence.
Usually costs: Half to one mark.

Exam tips

2024 Q8(i) and 2025 Q5(ii) both gave a factor with an unknown $k$ , then asked for complete factorisation; expected values were $k = 4$ and $k = 7$ .
Show $f (a)$ with every term written out; examiners award the substitution step.
When two unknowns are involved, number your equations (1) and (2) and show how you solve them.
In MCQs, test each option quickly; often only one gives zero.

Practice questions on Factor Theorem

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
If $(x - 2)$ is a factor of $x^{2} + k x + 6$ , then $k$ is:
(a) 5
(b) −5
(c) 3
(d) −3
Hint
$4 + 2 k + 6 = 0$ .
Answer
(b) −5
Q2 · 3 marks · Short answerBoard level
Is $(x + 1)$ a factor of $x^{3} + 3 x^{2} + 3 x + 1$ ? Justify.
Hint
Find $f (- 1)$ .
Answer
$f (- 1) = - 1 + 3 - 3 + 1 = 0$ , so yes.
Q3 · 3 marks · Short answerHarder board level
Find $k$ if $(3 x - 2)$ is a factor of $3 x^{3} + k x^{2} - 5 x + 2$ .
Hint
Substitute $x = \frac{2}{3}$ .
Answer
$\frac{8}{9} + \frac{4 k}{9} - \frac{10}{3} + 2 = \frac{4 k - 4}{9} = 0 \Rightarrow k = 1$ .
Q4 · 3 marks · Short answerHarder board level
$(x - 2)$ and $(x + 1)$ are factors of $2 x^{3} + a x^{2} + b x - 4$ . Find $a$ and $b$ .
Hint
Two equations: $2 a + b = - 6$ and $a - b = 6$ .
Answer
$a = 0$ , $b = - 6$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $(x - 3)$ is a factor of $x^{3} - 27$ .
Reason (R): If $f (a) = 0$ , then $(x - a)$ is a factor of $f (x)$ .
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
$3^{3} - 27 = 0$ .
Answer
(c)
Q6 · 4 marks · Short answerHarder board level
$(x - 2)$ is a factor of $f (x) = x^{3} + p x^{2} + q x + 6$ , and $f (x)$ leaves a remainder 6 when divided by $(x - 1)$ . Find $p$ and $q$ .
Hint
$f (2) = 0$ and $f (1) = 6$ .
Answer
$2 p + q = - 7$ and $p + q = - 1$ , so $p = - 6$ , $q = 5$ .
Q7 · 4 marks · Short answerHOTS / competency
Find $m$ if $x^{2} - 4$ is a factor of $x^{3} + m x^{2} - 4 x - 8$ , and factorise the cubic completely.
Hint
Both $(x - 2)$ and $(x + 2)$ must be factors.
Answer
$f (2) = 4 m - 8 = 0$ and $f (- 2) = 4 m - 8 = 0$ , so $m = 2$ . Then $x^{3} + 2 x^{2} - 4 x - 8 = (x - 2) (x + 2)^{2}$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the factor theorem.Show answer
$(x - a)$ is a factor of $f (x)$ if and only if $f (a) = 0$ .
2.What do you substitute to test the factor $(2 x + 3)$ ?Show answer
$x = - \frac{3}{2}$ .
3.Is $(x - 1)$ a factor of $x^{3} - 1$ ?Show answer
Yes, $1 - 1 = 0$ .
4. $(x - 1)$ is a factor of $x^{2} + x + k$ . Find $k$ .Show answer
$1 + 1 + k = 0 \Rightarrow k = - 2$ .
5. $f (- 3) = 5$ . Is $(x + 3)$ a factor?Show answer
No, the remainder is 5.

Factor Theorem: frequently asked questions

What is the factor theorem?

It says that (x − a) is a factor of a polynomial f(x) exactly when f(a) = 0. It is the remainder theorem in the special case where the remainder is zero.

How do I use the factor theorem for a factor like (2x − 3)?

Set 2x − 3 = 0 to get x = 3/2, and check whether f(3/2) = 0.

What degree of polynomial does ICSE use?

The CISCE syllabus says the polynomial will not exceed degree 3.