MathsCBSE Class 9 · Exploring Algebraic Identities

Factorising Using Algebraic Identities: Class 9

Updated 17 October 2026 · Fact-checked

To factorise, first look for a common factor, then match the expression to an identity: $a^2 \pm 2ab + b^2 = (a \pm b)^2$, $a^2 - b^2 = (a-b)(a+b)$, or the trinomial square. If none fits and the expression is $ax^2 + bx + c$, split the middle term using two numbers whose product is $ac$ and whose sum is $b$. The same identities also evaluate numbers like $998^2$ or $10.3 \times 9.7$ quickly.

Always remove a common factor first, then match the shape to an identity, and keep factorising until no factor can be broken further.

Illustration drawn for QuestionBuddy.

Understand Factorising Using Algebraic Identities

An identity is true for every value of the variables, unlike an equation, which holds only for particular values. Factorising reverses expansion: you recognise the product shape and write it as factors. Three recognisers do most of the work.

Perfect squares. $a^{2} + 2 ab + b^{2} = (a + b)^{2}$ and $a^{2} - 2 ab + b^{2} = (a - b)^{2}$ . An expression is a perfect square when two terms are perfect squares (such as $49 a^{2}$ and $16 b^{2}$ ) and the middle term is twice the product of their roots, so $49 a^{2} - 56 ab + 16 b^{2}$ has $2 \times 7 a \times 4 b = 56 ab$ and equals $(7 a - 4 b)^{2}$ . If the middle term does not match, it is not a perfect square, so do not force it.

Difference of squares. $a^{2} - b^{2} = (a - b) (a + b)$ . Look for two squares separated by a minus sign, including hidden ones such as $x^{4} - 81 = (x^{2} - 9) (x^{2} + 9) = (x - 3) (x + 3) (x^{2} + 9)$ . Keep factorising until no factor can be broken further over the rational numbers (here $x^{2} + 9$ stays). The same identity is the quick tool for arithmetic: $5 3^{2} - 4 7^{2} = (53 - 47) (53 + 47) = 600$ .

Square of a trinomial. $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 ab + 2 b c + 2 c a$ . It expands three-term brackets, and read backwards it factorises expressions with six terms of this shape. A powerful variant groups a perfect square and then subtracts a square: $9 x^{2} - 12 x + 4 - y^{2} = (3 x - 2)^{2} - y^{2} = (3 x - 2 - y) (3 x - 2 + y)$ .

Splitting the middle term. For $a x^{2} + b x + c$ , find two numbers $p, q$ with $p + q = b$ and $pq = a c$ . Then write $b x = p x + q x$ and factorise by grouping. For $12 x^{2} - 7 x y - 12 y^{2}$ : $a c = - 144$ , $b = - 7$ , so $p = 9$ and $q = - 16$ , giving $12 x^{2} + 9 x y - 16 x y - 12 y^{2} = 3 x (4 x + 3 y) - 4 y (4 x + 3 y) = (3 x - 4 y) (4 x + 3 y)$ . Check by expanding or by putting in numbers such as $x = 2$ , $y = 1$ : both sides equal $22$ .

Using identities for numbers. Write one number as a base plus or minus a small amount: $99 8^{2} = (1000 - 2)^{2} = 996004$ and $10.3 \times 9.7 = (10 + 0.3) (10 - 0.3) = 100 - 0.09 = 99.91$ . Squares of numbers ending in $5$ follow $(10 a + 5)^{2} = 100 a (a + 1) + 25$ .

Method to remember. (1) Take out the common factor. (2) Count terms: two terms suggest a difference of squares, three terms a perfect square or a split, four terms suggest grouping. (3) Factorise fully. (4) Verify by expanding mentally or substituting a number.

Factorising Using Algebraic Identities: formulas to remember

Square of a sum: $(a + b)^{2} = a^{2} + 2 ab + b^{2}$; Perfect square when the middle term is twice the product.
Square of a difference: $(a - b)^{2} = a^{2} - 2 ab + b^{2}$; Middle term negative.
Difference of squares: $a^{2} - b^{2} = (a - b) (a + b)$; Two squares with a minus sign.
Square of a trinomial: $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 ab + 2 b c + 2 c a$; Each cross-term is doubled.
Product of two binomials: $(x + p) (x + q) = x^{2} + (p + q) x + pq$; Reverse it to split a monic quadratic.
Splitting rule: $a x^{2} + b x + c : p + q = b, pq = a c$; Split $b x$ as $p x + q x$ and group.
Squares ending in 5: $(10 a + 5)^{2} = 100 a (a + 1) + 25$; Example: $6 5^{2} = 100 \cdot 6 \cdot 7 + 25 = 4225$ .

How to solve Factorising Using Algebraic Identities questions

Work in a fixed order so no factor is missed.

1Take out the greatest common factor of all terms first.
2Count the terms: two terms (look for $a^{2} - b^{2}$ ), three terms (perfect square or split the middle term), four or more (group or use the trinomial square).
3For a perfect square check the middle term equals $2 \times$ the product of the two square roots.
4For $a x^{2} + b x + c$ list pairs with product $a c$ until the sum is $b$ ; split $b x$ and group.
5Factorise each factor again if it is itself a difference of squares or can be split.
6Verify by expanding, or by substituting a small value such as $x = 2$ , and write the answer as a product.

Factorising Using Algebraic Identities 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.

Squares ending in 5

Suppose a tiling contractor must find the area of a square floor of side 75 feet and wants a quick check without a calculator.

Use an identity to find $7 5^{2}$ and explain why it works.

Try it first, then show the solutionSolution

Write $75 = 10 \times 7 + 5$ , so $a = 7$ in $(10 a + 5)^{2}$ .
$(10 a + 5)^{2} = 100 a^{2} + 100 a + 25 = 100 a (a + 1) + 25$ .
With $a = 7$ : $100 \times 7 \times 8 + 25 = 5600 + 25 = 5625$ .

Answer: The area is 5625 square feet.

Memory hook: a times (a + 1), then write 25 at the end.

Tricks to remember Factorising Using Algebraic Identities

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

Two squares, one minus
Spot $□^{2} - □^{2}$ and write (first minus second)(first plus second).
Use it when: Difference of squares, including numerical shortcuts.
Twice the product
Middle term is twice the product of the two roots, or the expression is not a perfect square.
Use it when: Testing perfect squares.
Product $a c$ , sum $b$
Multiply the outer coefficients, then hunt for a pair that adds to the middle one.
Use it when: Splitting the middle term.

Factorising Using Algebraic Identities: solved examples

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

Example 1: Perfect square · 3 marks · Short answerBoard level

Factorise $49 a^{2} - 56 ab + 16 b^{2}$ and verify the result by substituting $a = 1$ , $b = 1$ .

Try it first, then show the solutionSolution

$49 a^{2} = (7 a)^{2}$ and $16 b^{2} = (4 b)^{2}$ . (½)
Middle term: $2 \times 7 a \times 4 b = 56 ab$ , which matches with a minus sign. (1)
So the expression is $(7 a - 4 b)^{2}$ . (1)
Check: LHS $= 49 - 56 + 16 = 9$ and RHS $= (7 - 4)^{2} = 9$ . (½)

Answer: $49 a^{2} - 56 ab + 16 b^{2} = (7 a - 4 b)^{2}$ .

Example 2: Numerical shortcuts · 3 marks · Short answerBoard level

Using identities, evaluate (i) $100 4^{2}$ and (ii) $9.8 \times 10.2$ .

Try it first, then show the solutionSolution

(i) $1004 = 1000 + 4$ , so $100 4^{2} = 100 0^{2} + 2 (1000) (4) + 4^{2}$ . (1)
$= 1000000 + 8000 + 16 = 1008016$ . (½)
(ii) $9.8 = 10 - 0.2$ and $10.2 = 10 + 0.2$ , so the product is $1 0^{2} - 0. 2^{2}$ . (1)
$= 100 - 0.04 = 99.96$ . (½)

Answer: (i) $1008016$ ; (ii) $99.96$ .

Example 3: Split a two-variable quadratic · 5 marks · Long answerHarder board level

Factorise $12 x^{2} - 7 x y - 12 y^{2}$ , and verify at $x = 2$ , $y = 1$ .

Try it first, then show the solutionSolution

Here $a c = 12 \times (- 12) = - 144$ and $b = - 7$ . (1)
Find a pair: product $- 144$ , sum $- 7$ : the pair $9$ and $- 16$ works. (1)
$12 x^{2} + 9 x y - 16 x y - 12 y^{2}$ . (½)
$= 3 x (4 x + 3 y) - 4 y (4 x + 3 y)$ . (1)
$= (3 x - 4 y) (4 x + 3 y)$ . (½)
Check: LHS $= 48 - 14 - 12 = 22$ ; RHS $= (6 - 4) (8 + 3) = 2 \times 11 = 22$ . (1)

Answer: $12 x^{2} - 7 x y - 12 y^{2} = (3 x - 4 y) (4 x + 3 y)$ .

Example 4: Reciprocal sums · 3 marks · Short answerHarder board level

If $x + \frac{1}{x} = 7$ , find $x^{2} + \frac{1}{x ^{2}}$ , $x^{4} + \frac{1}{x ^{4}}$ and $(x - \frac{1}{x})^{2}$ .

Try it first, then show the solutionSolution

$(x + \frac{1}{x})^{2} = x^{2} + 2 + \frac{1}{x ^{2}}$ , so $x^{2} + \frac{1}{x ^{2}} = 49 - 2 = 47$ . (1)
Square again: $x^{4} + \frac{1}{x ^{4}} = 4 7^{2} - 2 = 2209 - 2 = 2207$ . (1)
$(x - \frac{1}{x})^{2} = x^{2} - 2 + \frac{1}{x ^{2}} = 47 - 2 = 45$ . (1)

Answer: $47$ , $2207$ and $45$ .

Example 5: Path around a lawn · 4 marks · Case-basedHOTS / competency

Case: a square lawn of side $s$ metres has a path of uniform width $w$ metres all around it. (i) Show that the area of the path is $4 w (s + w)$ . (ii) Find the area when $s = 47$ and $w = 3$ . (iii) A path 2 m wide has area 360 m $^{2}$ : find $s$ . (iv) Why is the identity better than expanding?

Try it first, then show the solutionSolution

(i) The outer square has side $s + 2 w$ , so path area $= (s + 2 w)^{2} - s^{2} = (s + 2 w - s) (s + 2 w + s) = 2 w (2 s + 2 w) = 4 w (s + w)$ . (1)
(ii) $4 \times 3 \times (47 + 3) = 12 \times 50 = 600$ m $^{2}$ ; check $5 3^{2} - 4 7^{2} = 2809 - 2209 = 600$ . (1)
(iii) $4 \times 2 \times (s + 2) = 360$ , so $s + 2 = 45$ and $s = 43$ m; check $4 7^{2} - 4 3^{2} = 2209 - 1849 = 360$ . (1)
(iv) The difference of squares gives a short product of two small numbers and works for any $s$ and $w$ , whereas expanding the squares needs big numbers. (1)

Answer: (i) shown; (ii) 600 m $^{2}$ ; (iii) $s = 43$ m; (iv) the factored form is quicker and general.

Where students lose marks in Factorising Using Algebraic Identities

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

Mistake: Forcing a perfect square
Matching the end terms and ignoring the middle term.
Wrong working
$4 x^{2} + 10 x + 9 = (2 x + 3)^{2}$ .
Right working
$(2 x + 3)^{2} = 4 x^{2} + 12 x + 9$ , so the middle term should be $12 x$ ; the given expression is not a perfect square.
Do this: Compute twice the product of the roots and compare.
Usually costs: The whole answer.
Mistake: Stopping before fully factorised
A factor like $x^{2} - 9$ is left unbroken.
Wrong working
$x^{4} - 81 = (x^{2} - 9) (x^{2} + 9)$ .
Right working
$x^{4} - 81 = (x - 3) (x + 3) (x^{2} + 9)$ .
Do this: Re-inspect every factor.
Usually costs: 1 mark.
Mistake: Wrong signs in the split
Choosing a pair with the right product but the wrong sum sign.
Wrong working
$6 x^{2} - 5 x - 6$ split as $+ 9 x - 4 x$ .
Right working
The pair has sum $- 5$ and product $- 36$ : $- 9$ and $+ 4$ .
Do this: Check both product and sum before grouping.
Usually costs: 2 marks.

Exam tips

Always begin with the common factor; it is the most missed step.
Show the identity name or form once, then apply it; examiners award a method mark for it.
Verify with a quick substitution when the expression has two variables.
In "evaluate" questions, show the decomposition (for example $1000 - 2$ ) rather than just the answer.

Practice questions on Factorising Using Algebraic Identities

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 $x^{2} + k x + 49$ is a perfect square and $k > 0$ , then $k$ equals
(a) 7
(b) 14
(c) 49
(d) 98
Hint
The middle term must be twice the product of $x$ and $7$ .
Answer
(b) 14, since $x^{2} + 14 x + 49 = (x + 7)^{2}$ and $2 \times x \times 7 = 14 x$ .
Q2 · 3 marks · Short answerBoard level
Expand $(3 a - 2 b + c)^{2}$ .
Hint
Use the trinomial identity; take care with the signs of cross-terms.
Answer
$9 a^{2} + 4 b^{2} + c^{2} + 2 (3 a) (- 2 b) + 2 (- 2 b) (c) + 2 (3 a) (c) = 9 a^{2} + 4 b^{2} + c^{2} - 12 ab - 4 b c + 6 a c$ .
Q3 · 3 marks · Short answerHarder board level
Factorise completely $9 x^{2} - 12 x + 4 - y^{2}$ .
Hint
The first three terms form a perfect square.
Answer
$9 x^{2} - 12 x + 4 = (3 x - 2)^{2}$ , so the expression is $(3 x - 2)^{2} - y^{2} = (3 x - 2 - y) (3 x - 2 + y)$ .
Q4 · 5 marks · Long answerHarder board level
If $a + b + c = 10$ and $a^{2} + b^{2} + c^{2} = 38$ , find $ab + b c + c a$ and $(a - b)^{2} + (b - c)^{2} + (c - a)^{2}$ . Verify that $(a + b + c)^{2} \geq 3 (ab + b c + c a)$ holds here.
Hint
Square $a + b + c$ ; expand each square in the second sum.
Answer
$(a + b + c)^{2} = 100 = 38 + 2 (ab + b c + c a)$ , so $ab + b c + c a = 31$ . $(a - b)^{2} + (b - c)^{2} + (c - a)^{2} = 2 (a^{2} + b^{2} + c^{2}) - 2 (ab + b c + c a) = 76 - 62 = 14$ . Since this sum of squares is at least 0, $(a + b + c)^{2} \geq 3 (ab + b c + c a)$ ; numerically $100 \geq 93$ is true.
Q5 · 4 marks · Case-basedHOTS / competency
Case: a rectangular hall has floor area $(6 x^{2} - 5 x - 6)$ square metres. (i) Factorise the area to find the two sides. (ii) If $x = 3$ , find the sides and the perimeter. (iii) For which values of $x$ are both sides positive? (iv) Check your factors by multiplying them back.
Hint
Split $- 5 x$ using the pair with product $- 36$ .
Answer
(i) $6 x^{2} - 9 x + 4 x - 6 = 3 x (2 x - 3) + 2 (2 x - 3) = (3 x + 2) (2 x - 3)$ . (ii) Sides $11$ m and $3$ m (area $33$ m $^{2}$ ), perimeter $2 (11 + 3) = 28$ m. (iii) $3 x + 2 > 0$ and $2 x - 3 > 0$ need $x > 1.5$ . (iv) $(3 x + 2) (2 x - 3) = 6 x^{2} - 9 x + 4 x - 6 = 6 x^{2} - 5 x - 6$ .

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. $a^{2} - b^{2}$ equals?Show answer
$(a - b) (a + b)$ .
2.Middle term of a perfect square?Show answer
Twice the product of the two roots.
3.Splitting rule for $a x^{2} + b x + c$ ?Show answer
Pair with product $a c$ and sum $b$ .
4. $(a + b + c)^{2}$ has how many cross-terms?Show answer
Three, each doubled: $2 ab, 2 b c, 2 c a$ .
5. $(10 a + 5)^{2}$ equals?Show answer
$100 a (a + 1) + 25$ .

Factorising Using Algebraic Identities: frequently asked questions

What is the difference between an identity and an equation?

An identity is true for all values of the variables, while an equation is true only for particular values, which you solve for.

Can every quadratic be factorised by splitting?

Only when suitable integer or simple rational pairs exist. Others, like x^2 + x + 1, do not factorise over the real numbers.

Is $x^2 + 9$ a difference of squares?

No. It is a sum of squares and does not factorise over the real numbers.