MathsICSE Class 10 · Factorisation of Polynomials

Factorising a Cubic Polynomial Completely: Method and Solved Questions (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

To factorise a cubic completely, find one value a with f(a) = 0 by trial (divisors of the constant term, or fractions when the leading coefficient is not 1), so (x − a) is a factor. Divide f(x) by (x − a) to get a quadratic, factorise that, and write f(x) as a product of three linear factors.

Find one root, divide to get a quadratic, factorise it, then write all three linear factors as one product.

Illustration drawn for QuestionBuddy.

Understand Factorising a Cubic Completely

The CISCE syllabus asks you to factorise a polynomial (of degree at most 3) completely after obtaining one factor by the factor theorem. A typical board question has three steps: find a missing coefficient, show a factor, then 'hence factorise completely', as in 2024 Q8(i) and 2025 Q5(ii).

Step 1: find one factor. If no factor is given, try simple values. For $x^{3} + b x^{2} + c x + d$ , any integer root divides $d$ , so try $\pm 1, \pm 2, \pm 3, \dots$ among the divisors of $d$ . Two quick checks help: $f (1)$ is the sum of the coefficients, and $f (- 1)$ is the sum with the signs of odd-power terms changed. For a leading coefficient $p \neq = 1$ , also try fractions $\frac{q}{p}$ where $q$ divides the constant term.

Step 2: divide. Divide $f (x)$ by $(x - a)$ to get a quadratic quotient. Use long division, synthetic division, or comparing coefficients: write $f (x) = (x - a) (p x^{2} + q x + r)$ , read $p$ from the $x^{3}$ term and $r$ from the constant term, and find $q$ from the $x^{2}$ term. The examiners reported errors in dividing by $(x - 1)$ in 2024 Q8(i); synthetic division (bring down, multiply, add) avoids most of them.

Step 3: factorise the quadratic. Split the middle term. If the quadratic has no rational factors, leave it as it is (in ICSE board questions it nearly always factorises).

Step 4: write the product. The final answer must be a product: for example $(x - 1) (x + 4) (3 x - 1)$ . The examiners reported for both 2024 and 2025 candidates who found the factors but did not write the answer as a product of three factors.

Grouping shortcut. Some cubics factorise by grouping without any division: $x^{3} - 3 x^{2} - 4 x + 12 = x^{2} (x - 3) - 4 (x - 3) = (x - 3) (x^{2} - 4) = (x - 3) (x - 2) (x + 2)$ . Check whether the first two and last two terms share a bracket before dividing.

Check. Expand two of the factors and multiply by the third, or test one more value: a correct factorisation gives the same value as $f (x)$ at, say, $x = 2$ .

Factorising a Cubic Completely: formulas to remember

Factorised form of a cubic: $f (x) = (x - a) (p x^{2} + q x + r)$; Compare coefficients to find $p$ , $q$ , $r$ .
Trial values: $x = \pm \frac{divisor of constant term}{divisor of leading coefficient}$; Integer trials suffice when the leading coefficient is 1.
Quick tests: $f (1) = sum of coefficients$; If it is 0, $(x - 1)$ is a factor.
Synthetic division by a: $b_{0} = a_{0}, b_{k} = a_{k} + a b_{k - 1}$; Last value is the remainder (must be 0).
Grouping: $x^{2} (x - k) - m (x - k) = (x - k) (x^{2} - m)$; When the pairs share a bracket.

How to solve Factorising a Cubic Completely questions

Find one factor, divide, factorise the quadratic, write the product.

1If a coefficient is unknown, use the given factor to find it first.
2Find a root $a$ : try $f (1)$ , $f (- 1)$ , then other divisors of the constant term (and fractions if the leading coefficient is not 1).
3Divide by $(x - a)$ using synthetic division or comparing coefficients; the remainder must be 0.
4Factorise the quadratic quotient by splitting the middle term.
5Write $f (x)$ as a product of linear factors.
6Check by expanding or by testing one value.

Factorising a Cubic Completely 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.

Dimensions from a volume

Suppose a storage box has volume $x^{3} + 9 x^{2} + 26 x + 24$ cm³, and its edges are linear expressions in $x$ .

Find the edges, and the actual edges when $x = 6$ .

Try it first, then show the solutionSolution

$f (- 2) = - 8 + 36 - 52 + 24 = 0$ , so $(x + 2)$ is a factor.
$f (x) = (x + 2) (x^{2} + 7 x + 12) = (x + 2) (x + 3) (x + 4)$ .
At $x = 6$ : 8, 9 and 10 cm (volume 720 cm³).

Answer: $(x + 2)$ , $(x + 3)$ , $(x + 4)$ cm; 8 cm, 9 cm and 10 cm when $x = 6$ .

Memory hook: A cubic volume hides three edges; one factor reveals all three.

A cubic as three consecutive numbers

Imagine a puzzle: from an $n \times n \times n$ cube of small blocks you remove one row of $n$ blocks, leaving $n^{3} - n$ blocks.

Show that $n^{3} - n$ is the product of three consecutive whole numbers, and find it for $n = 5$ .

Try it first, then show the solutionSolution

$n^{3} - n = n (n^{2} - 1) = n (n - 1) (n + 1) = (n - 1) n (n + 1)$ .
For $n = 5$ : $4 \times 5 \times 6 = 120$ (and $125 - 5 = 120$ ).

Answer: $(n - 1) n (n + 1)$ ; 120 blocks for $n = 5$ .

Memory hook: Take out the common factor before anything else.

Tricks to remember Factorising a Cubic Completely

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

Find one, divide, split, multiply
F-D-S-M: Find a root, Divide, Split the quadratic, Multiply out as a product.
Use it when: Every 'factorise completely' question.
Sum of coefficients test
If the coefficients add to 0, $(x - 1)$ is a factor at once.
Use it when: Fast first trial.
Three brackets in the answer
A cubic that factorises completely ends as three brackets. If you have two, keep going.
Use it when: Avoiding the error in 2024 Q8(i) and 2025 Q5(ii).

Factorising a Cubic Completely: solved examples

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

Example 1: Recognising the full factorisation · 1 mark · MCQBoard level

$x^{3} - 7 x + 6$ factorises completely as:
(a) $(x - 1) (x - 2) (x + 3)$
(b) $(x + 1) (x + 2) (x - 3)$
(c) $(x - 1) (x + 2) (x - 3)$
(d) $(x - 1)^{2} (x + 6)$

Try it first, then show the solutionSolution

$f (1) = 1 - 7 + 6 = 0$ ; dividing gives $x^{2} + x - 6 = (x - 2) (x + 3)$ .

Answer: (a)

Example 2: Leading coefficient 1 · 3 marks · Short answerBoard level

Factorise completely: $x^{3} - 2 x^{2} - x + 2$ .

Try it first, then show the solutionSolution

$f (1) = 1 - 2 - 1 + 2 = 0$ , so $(x - 1)$ is a factor.
Synthetic division by 1: coefficients $1, - 2, - 1, 2 \to 1, - 1, - 2$ , remainder 0; quotient $x^{2} - x - 2$ .
$x^{2} - x - 2 = (x - 2) (x + 1)$ ; so $f (x) = (x - 1) (x - 2) (x + 1)$ .

Answer: $(x - 1) (x - 2) (x + 1)$

Example 3: Leading coefficient 2 · 4 marks · Short answerHarder board level

Factorise completely: $2 x^{3} + 3 x^{2} - 11 x - 6$ .

Try it first, then show the solutionSolution

$f (1) = - 12$ , $f (- 1) = 6$ , $f (2) = 16 + 12 - 22 - 6 = 0$ : $(x - 2)$ is a factor.
Synthetic division by 2: $2, 3, - 11, - 6 \to 2, 7, 3$ , remainder 0; quotient $2 x^{2} + 7 x + 3$ .
$2 x^{2} + 7 x + 3 = 2 x^{2} + 6 x + x + 3 = (2 x + 1) (x + 3)$ .
$f (x) = (x - 2) (2 x + 1) (x + 3)$ .

Answer: $(x - 2) (2 x + 1) (x + 3)$

Example 4: Find k, then factorise · 4 marks · Short answerHarder board level

$(x - 2)$ is a factor of $2 x^{3} + k x^{2} - 13 x + 6$ . Find $k$ and factorise the polynomial completely.

Try it first, then show the solutionSolution

$f (2) = 16 + 4 k - 26 + 6 = 4 k - 4 = 0 \Rightarrow k = 1$ .
$2 x^{3} + x^{2} - 13 x + 6 = (x - 2) (2 x^{2} + 5 x - 3)$ (synthetic division: $2, 1, - 13, 6 \to 2, 5, - 3$ , remainder 0).
$2 x^{2} + 5 x - 3 = (2 x - 1) (x + 3)$ .
$f (x) = (x - 2) (2 x - 1) (x + 3)$ .

Answer: $k = 1$ ; $(x - 2) (2 x - 1) (x + 3)$

Example 5: No integer root at all · 4 marks · Short answerHOTS / competency

Factorise completely: $12 x^{3} + 16 x^{2} - 5 x - 3$ .

Try it first, then show the solutionSolution

Integer trials fail ( $f (1) = 20$ , $f (- 1) = 6$ ); try fractions $\frac{q}{p}$ with $q ∣ 3$ , $p ∣ 12$ .
$f (\frac{1}{2}) = \frac{12}{8} + 4 - \frac{5}{2} - 3 = 0$ , so $(2 x - 1)$ is a factor.
$12 x^{3} + 16 x^{2} - 5 x - 3 = (2 x - 1) (6 x^{2} + 11 x + 3)$ (compare coefficients: $6 x^{2}$ , constant $3$ , middle $2 \cdot q - 6 = 16 \Rightarrow q = 11$ ).
$6 x^{2} + 11 x + 3 = (3 x + 1) (2 x + 3)$ ; so $f (x) = (2 x - 1) (3 x + 1) (2 x + 3)$ .

Answer: $(2 x - 1) (3 x + 1) (2 x + 3)$

Example 6: Case study: a cuboid's volume · 4 marks · Case-basedHOTS / competency

A cuboid has volume $V (x) = 4 x^{3} + 8 x^{2} - x - 2$ cubic units.
(a) Show that $(x + 2)$ is a factor of $V (x)$ .
(b) Factorise $V (x)$ completely to find expressions for the edges.
(c) Find the edges and the volume when $x = 3$ .

Try it first, then show the solutionSolution

(a) $V (- 2) = - 32 + 32 + 2 - 2 = 0$ .
(b) Grouping: $4 x^{2} (x + 2) - 1 (x + 2) = (x + 2) (4 x^{2} - 1) = (x + 2) (2 x - 1) (2 x + 1)$ .
(c) $x = 3$ : edges 5, 5, 7; volume $= 175$ (check: $108 + 72 - 3 - 2 = 175$ ).

Answer: (a) $V (- 2) = 0$ (b) $(x + 2) (2 x - 1) (2 x + 1)$ (c) 5, 5, 7 units; 175 cubic units.

Where students lose marks in Factorising a Cubic Completely

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

Mistake: Stopping at a linear × quadratic
'Completely' means every factor that can be split must be split.
Wrong working
$(x - 2) (2 x^{2} - 3 x + 1)$ .
Right working
$(x - 2) (x - 1) (2 x - 1)$ .
Do this: Always try to factorise the quadratic.
Usually costs: Reported for 2024 Q8(i) and 2025 Q5(ii).
Mistake: Errors in long division
Subtracting rows with negatives is error-prone.
Wrong working
Quotient $3 x^{2} + 5 x - 4$ for $3 x^{3} + 8 x^{2} - 15 x + 4 \div (x - 1)$ .
Right working
Synthetic division: $3, 8, - 15, 4 \to 3, 11, - 4$ ; quotient $3 x^{2} + 11 x - 4$ .
Do this: Use synthetic division and check the remainder is 0.
Usually costs: Reported for 2024 Q8(i).
Mistake: Not writing the answer as a product
Listing roots is not a factorisation.
Wrong working
$x = 2, 1, \frac{1}{2}$ .
Right working
$(x - 2) (x - 1) (2 x - 1)$ .
Do this: End with 'f(x) = (…)(…)(…)'.
Usually costs: The final mark; reported for 2025 Q5(ii).

Exam tips

In 2024 Q8(i) the expected answer was $(x - 1) (x + 4) (3 x - 1)$ after finding $k = 4$ ; in 2025 Q5(ii) it was $(x - 2) (x - 1) (2 x - 1)$ after $k = 7$ .
Show the division (or comparing coefficients) clearly; the quotient carries a mark.
Before dividing, look for a grouping that works; it saves time.
Check by expanding two brackets and multiplying by the third.

Practice questions on Factorising a Cubic Completely

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
$x^{3} + 2 x^{2} - x - 2$ factorises completely as:
(a) $(x + 2) (x - 1) (x + 1)$
(b) $(x - 2) (x - 1) (x + 1)$
(c) $(x + 2) (x - 1)^{2}$
(d) $(x + 2) (x^{2} + 1)$
Hint
Group: $x^{2} (x + 2) - (x + 2)$ .
Answer
(a)
Q2 · 3 marks · Short answerBoard level
Factorise completely: $x^{3} + 2 x^{2} - 9 x - 18$ .
Hint
Grouping works.
Answer
$x^{2} (x + 2) - 9 (x + 2) = (x + 2) (x^{2} - 9) = (x + 2) (x - 3) (x + 3)$ .
Q3 · 3 marks · Short answerHarder board level
Factorise completely: $2 x^{3} - x^{2} - 13 x - 6$ .
Hint
Try $x = - 2$ .
Answer
$f (- 2) = 0$ ; $(x + 2) (2 x^{2} - 5 x - 3) = (x + 2) (2 x + 1) (x - 3)$ .
Q4 · 4 marks · Short answerHarder board level
$(x + 1)$ is a factor of $x^{3} + a x^{2} - x - 6$ . Find $a$ and factorise completely.
Hint
$f (- 1) = - 1 + a + 1 - 6$ .
Answer
$a = 6$ ; $x^{3} + 6 x^{2} - x - 6 = (x + 1) (x^{2} + 5 x - 6) = (x + 1) (x + 6) (x - 1)$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $x^{3} + 3 x^{2} + 3 x + 1 = (x + 1)^{3}$ .
Reason (R): $f (- 1) = 0$ for $f (x) = x^{3} + 3 x^{2} + 3 x + 1$ .
(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
Does $f (- 1) = 0$ show that $(x + 1)$ appears three times?
Answer
(d). Both are true, but $f (- 1) = 0$ only shows $(x + 1)$ is a factor; (A) needs the full expansion of $(x + 1)^{3}$ .
Q6 · 4 marks · Short answerHarder board level
Factorise completely: $6 x^{3} + 13 x^{2} + x - 2$ .
Hint
Try $x = - 2$ .
Answer
$f (- 2) = 0$ ; $(x + 2) (6 x^{2} + x - 1) = (x + 2) (3 x - 1) (2 x + 1)$ .
Q7 · 4 marks · Short answerHOTS / competency
Find the quadratic factor common to $x^{3} - 3 x^{2} - x + 3$ and $x^{3} + 2 x^{2} - x - 2$ .
Hint
Factorise each completely by grouping.
Answer
$(x - 3) (x - 1) (x + 1)$ and $(x + 2) (x - 1) (x + 1)$ ; common factor $(x - 1) (x + 1) = x^{2} - 1$ .

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.Factorise $x^{3} - x$ .Show answer
$x (x - 1) (x + 1)$ .
2.Coefficients of $x^{3} - 4 x^{2} + x + 2$ add to 0. Which factor does that give?Show answer
$(x - 1)$ .
3.Synthetic division of $x^{3} - 4 x^{2} + x + 2$ by 1 gives which quotient?Show answer
$x^{2} - 3 x - 2$ .
4.Which fractions should you try for $2 x^{3} + \dots + 3$ ?Show answer
$\pm \frac{1}{2}$ , $\pm \frac{3}{2}$ (as well as $\pm 1, \pm 3$ ).
5.Factorise $x^{2} (x + 5) - 9 (x + 5)$ .Show answer
$(x + 5) (x - 3) (x + 3)$ .

Factorising a Cubic Completely: frequently asked questions

How do I factorise a cubic polynomial completely?

Find one root a by trial so that (x − a) is a factor, divide the cubic by (x − a) to get a quadratic, factorise the quadratic, and write the cubic as a product of the three factors.

What values should I try to find the first factor?

Divisors of the constant term, positive and negative. If the leading coefficient is not 1, also try fractions with the divisors of the leading coefficient in the denominator.

Do I have to use long division?

No. Synthetic division or comparing coefficients is acceptable and usually quicker, as long as your working is shown.