MathsICSE Class 9 · Factorisation

Splitting the Middle Term and Factorisation by Substitution (ICSE Class 9)

Q: What if no pair works?

Then the quadratic does not factorise over the integers.

Q: Can I use the quadratic formula instead?

Class IX asks for factorisation by splitting; the formula comes later.

Q: Does the order of p and q matter?

No; the grouping gives the same factors in a different order.

Updated 17 October 2026 · Fact-checked

To factorise ax² + bx + c, find two numbers whose product is ac and whose sum is b, split bx into those two terms and group. If an expression repeats a block, such as (x² + x), replace the block by a single letter, factorise, and substitute back.

Find two numbers with product ac and sum b, split the middle term, group so the same bracket appears twice, then factor it out.

Illustration drawn for QuestionBuddy.

Understand Splitting the Middle Term and Factorising by Substitution

Splitting the middle term. For $a x^{2} + b x + c$ find two numbers $p, q$ with
$ $p + q = b, pq = a c .$ $
Then $a x^{2} + b x + c = a x^{2} + p x + q x + c$ , and group.

*Example.* $6 x^{2} + 11 x - 10$ : $a c = - 60$ , $b = 11$ , so $p, q = 15, - 4$ (product $- 60$ , sum $11$ ).
$6 x^{2} + 15 x - 4 x - 10 = 3 x (2 x + 5) - 2 (2 x + 5) = (3 x - 2) (2 x + 5)$ .

Using the signs. If $a c > 0$ , $p$ and $q$ have the same sign as $b$ . If $a c < 0$ , they have opposite signs and the larger (in size) one has the sign of $b$ . List factor pairs of $∣ a c ∣$ systematically.

Check. Multiply the two binomials: the outer and inner products must add to the middle term.

Substitution. When the same block repeats, set it equal to a new letter. For $(x^{2} + x)^{2} - 8 (x^{2} + x) + 12$ let $y = x^{2} + x$ :
$y^{2} - 8 y + 12 = (y - 2) (y - 6)$ , so the expression is $(x^{2} + x - 2) (x^{2} + x - 6)$ . Each of these splits again: $(x + 2) (x - 1) (x + 3) (x - 2)$ . Always substitute back and factorise completely.

For $2 (a + b)^{2} - 5 (a + b) - 12$ let $y = a + b$ : $2 y^{2} - 5 y - 12 = (2 y + 3) (y - 4)$ , giving $(2 a + 2 b + 3) (a + b - 4)$ .

A completing-the-square trick. $x^{4} + 4 = (x^{2} + 2)^{2} - 4 x^{2} = (x^{2} - 2 x + 2) (x^{2} + 2 x + 2)$ . It is an add-and-subtract idea that turns a sum into a difference of squares.

Level. Class IX is a school-exam class. Splitting is routine; the substitution and add-and-subtract items are at the top of the school range and in foundation-olympiad territory.

Splitting the Middle Term and Factorising by Substitution: formulas to remember

Split condition: $p + q = b, pq = a c$; For ax² + bx + c.
Monic quadratic: $x^{2} + (p + q) x + pq = (x + p) (x + q)$; When a = 1.
Substitution: $y = repeated block$; Factorise in y, then substitute back.
Add and subtract: $x^{4} + 4 = (x^{2} + 2)^{2} - (2 x)^{2}$; Creates a difference of squares.

How to solve Splitting the Middle Term and Factorising by Substitution questions

Find p and q from the product ac and sum b, split, group, check.

1Compute $a c$ and write the factor pairs.
2Pick the pair whose sum is $b$ (watch the signs).
3Split the middle term and group in pairs.
4Take out the common bracket.
5For a repeated block, set $y$ = the block, factorise, substitute back.
6Factorise each resulting factor completely; multiply to check.

Splitting the Middle Term and Factorising by Substitution 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 of a rectangular garden

Suppose a rectangular garden has area $(6 x^{2} + 7 x - 3)$ m² and its sides are linear expressions in $x$ .

Find the sides, and the perimeter when $x = 2$ .

Try it first, then show the solutionSolution

$a c = - 18$ , $b = 7$ : $9, - 2$ . $6 x^{2} + 9 x - 2 x - 3 = 3 x (2 x + 3) - (2 x + 3) = (3 x - 1) (2 x + 3)$ .
At $x = 2$ : sides $5$ m and $7$ m; area $35$ m² (check $24 + 14 - 3 = 35$ ); perimeter $2 (5 + 7) = 24$ m.

Answer: Sides $(3 x - 1)$ and $(2 x + 3)$ m; perimeter 24 m at $x = 2$ .

Memory hook: The product of the sides is the area.

Revenue as a product

Imagine a stall whose daily revenue in rupees is modelled as $x^{2} - 5 x + 6$ for $x$ boxes sold, written as price times number.

Factorise to express it as a product.

Try it first, then show the solutionSolution

Numbers with product 6 and sum $- 5$ : $- 2, - 3$ .
$x^{2} - 5 x + 6 = (x - 2) (x - 3)$ .

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

Memory hook: Sum is the middle, product is the constant.

Tricks to remember Splitting the Middle Term and Factorising by Substitution

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

Product to ac, sum to b
Look for p, q with product ac and sum b.
Use it when: Every splitting question.
Same bracket twice
After grouping, both groups must show the same bracket.
Use it when: Checking the split.
Repeated block, new letter
Replace a repeated expression with y.
Use it when: Substitution questions.

Splitting the Middle Term and Factorising by Substitution: solved examples

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

Example 1: Monic quadratic · 1 mark · MCQBoard level

$x^{2} + 7 x + 12$ factorises as:
(a) $(x + 2) (x + 6)$
(b) $(x + 3) (x + 4)$
(c) $(x - 3) (x - 4)$
(d) $(x + 1) (x + 12)$

Try it first, then show the solutionSolution

Product 12, sum 7: $3$ and $4$ .

Answer: (b)

Example 2: Splitting with a leading coefficient · 3 marks · Short answerBoard level

Factorise $6 x^{2} + 11 x - 10$ .

Try it first, then show the solutionSolution

$a c = - 60$ , $b = 11$ : $15$ and $- 4$ .
$6 x^{2} + 15 x - 4 x - 10 = 3 x (2 x + 5) - 2 (2 x + 5)$ .
$= (3 x - 2) (2 x + 5)$ .

Answer: $(3 x - 2) (2 x + 5)$

Example 3: Larger numbers · 4 marks · Short answerHarder board level

Factorise $12 x^{2} - x - 35$ .

Try it first, then show the solutionSolution

$a c = - 420$ , $b = - 1$ : need numbers differing by 1 with product 420 in size: $20$ and $21$ , so $20$ and $- 21$ .
$12 x^{2} + 20 x - 21 x - 35 = 4 x (3 x + 5) - 7 (3 x + 5)$ .
$= (4 x - 7) (3 x + 5)$ .

Answer: $(4 x - 7) (3 x + 5)$

Example 4: Substitution · 4 marks · Short answerHarder board level

Factorise completely $(x^{2} + x)^{2} - 8 (x^{2} + x) + 12$ .

Try it first, then show the solutionSolution

Let $y = x^{2} + x$ : $y^{2} - 8 y + 12 = (y - 2) (y - 6)$ .
$(x^{2} + x - 2) (x^{2} + x - 6)$ .
$x^{2} + x - 2 = (x + 2) (x - 1)$ and $x^{2} + x - 6 = (x + 3) (x - 2)$ .

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

Example 5: Add and subtract · 4 marks · Long answerHOTS / competency

Factorise $x^{4} + 4$ by writing it as a difference of two squares. Verify by checking the value at $x = 1$ .

Try it first, then show the solutionSolution

$x^{4} + 4 = x^{4} + 4 x^{2} + 4 - 4 x^{2} = (x^{2} + 2)^{2} - (2 x)^{2}$ .
$= (x^{2} + 2 - 2 x) (x^{2} + 2 + 2 x) = (x^{2} - 2 x + 2) (x^{2} + 2 x + 2)$ .
At $x = 1$ : LHS $= 5$ ; RHS $= (1) (5) = 5$ . ✓

Answer: $(x^{2} - 2 x + 2) (x^{2} + 2 x + 2)$

Where students lose marks in Splitting the Middle Term and Factorising by Substitution

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

Mistake: Wrong sign choice for p and q
The pair must multiply to ac and add to b with the correct signs.
Wrong working
For $x^{2} - 5 x + 6$ choose $2$ and $3$ .
Right working
Choose $- 2$ and $- 3$ .
Do this: Check the sum matches b.
Usually costs: Whole answer.
Mistake: Splitting using the constant only
For a ≠ 1, the product is ac.
Wrong working
$6 x^{2} + 11 x - 10$ : product $- 10$ .
Right working
Product $a c = - 60$ .
Do this: Multiply a and c.
Usually costs: Whole answer.
Mistake: Not substituting back
The final answer must be in the original variable.
Wrong working
$(y - 2) (y - 6)$ .
Right working
$(x^{2} + x - 2) (x^{2} + x - 6)$ , then split fully.
Do this: Always replace y.
Usually costs: Final mark.

Exam tips

Class IX is examined in school papers; show the split step, because the method mark depends on it.
Write the factor pairs of ac in a short list; do not guess.
After substituting back, factorise each new bracket completely.
The add-and-subtract trick is an extension for strong students; do not use it where splitting works.

Practice questions on Splitting the Middle Term and Factorising by Substitution

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
The factors of $x^{2} - 5 x + 6$ are:
(a) $(x + 2) (x + 3)$
(b) $(x - 2) (x - 3)$
(c) $(x - 1) (x - 6)$
(d) $(x + 1) (x - 6)$
Hint
Product 6, sum −5.
Answer
(b)
Q2 · 3 marks · Short answerHarder board level
Factorise $2 x^{2} + 7 x - 15$ .
Hint
$a c = - 30$ ; $10$ and $- 3$ .
Answer
$2 x^{2} + 10 x - 3 x - 15 = 2 x (x + 5) - 3 (x + 5) = (2 x - 3) (x + 5)$ .
Q3 · 4 marks · Short answerHarder board level
Factorise $2 (a + b)^{2} - 5 (a + b) - 12$ .
Hint
Let $y = a + b$ .
Answer
$2 y^{2} - 5 y - 12 = (2 y + 3) (y - 4)$ , so $(2 a + 2 b + 3) (a + b - 4)$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $x^{2} + 5 x + 6$ can be written as $x^{2} + 2 x + 3 x + 6$ .
Reason (R): $2$ and $3$ have sum $5$ and product $6$ .
(a) Both (A) and (R) are true and (R) explains (A).
(b) Both are true but (R) does not explain (A).
(c) (A) is true, (R) is false.
(d) (A) is false, (R) is true.
Hint
Check the split rule.
Answer
(a)
Q5 · 4 marks · Case-basedHOTS / competency
Case study: A rectangular plot has area $(6 x^{2} + 7 x - 3)$ m².
(a) Factorise the area expression to find the length and breadth.
(b) Find the dimensions when $x = 2$ .
(c) Find the area and perimeter when $x = 2$ .
Hint
Find two numbers with product $- 18$ and sum $7$ .
Answer
(a) $(3 x - 1)$ and $(2 x + 3)$ (b) $5$ m and $7$ m (c) area $35$ m², perimeter $24$ m.

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.What two conditions define p and q in splitting?Show answer
p + q = b and pq = ac.
2.Factorise x² − 5x + 6.Show answer
(x − 2)(x − 3).
3.When do you use substitution?Show answer
When a block of the expression repeats.
4.After substitution, what must you do?Show answer
Replace the new letter and factorise further.

Splitting the Middle Term and Factorising by Substitution: frequently asked questions

What if no pair works?

Then the quadratic does not factorise over the integers.

Can I use the quadratic formula instead?

Class IX asks for factorisation by splitting; the formula comes later.

Does the order of p and q matter?

No; the grouping gives the same factors in a different order.