Factorising with identities

Some expressions look impossible to factorise by ordinary common-factor grouping — until you recognise them as an algebraic identity wearing a disguise. $50x^2 - 8y^2$ looks messy, but strip out a common factor of $2$ and it becomes $2(25x^2 - 4y^2)$ , an instantly recognisable difference of squares. This is the second technique inside the Class 8 Factorisation chapter, and it runs the Algebraic Identities you already expanded — squares of sums and differences, and sum/difference of cubes — in reverse. Instead of multiplying $(a+b)(a-b)$ out to $a^2-b^2$ , you start from $a^2-b^2$ and work backwards to $(a+b)(a-b)$ . The entire skill is pattern-matching: spotting perfect squares and cubes fast, checking whether the middle term fits, and pulling out any common factor first so the pattern becomes visible.

The core insight

Factorising by identities means matching an expanded expression against a known identity — difference of squares, perfect-square trinomial, or sum/difference of cubes — and writing the corresponding factored form directly, without long expansion.

The core idea

Four identities, used in reverse. The same identities you learned for expansion now run backwards to factorise:

a^2 - b^2 = (a-b)(a+b)

where: $a^2$ and $b^2$ are the two perfect-square terms, joined by a minus sign.

a^2 + 2ab + b^2 = (a+b)^2, \qquad a^2 - 2ab + b^2 = (a-b)^2

where: the first and last terms are perfect squares, and the middle term must equal exactly $\pm 2ab$ — twice the product of the square roots of the outer terms.

a^3 + b^3 = (a+b)(a^2-ab+b^2), \qquad a^3 - b^3 = (a-b)(a^2+ab+b^2)

where: $a$ and $b$ are the cube roots of the two terms; the second bracket always carries a $+ab$ middle term, regardless of whether the cubes are added or subtracted.

Recognising the pattern. For a difference of squares, the expression must be exactly two terms, both perfect squares, separated by a minus sign. For a perfect-square trinomial, there must be three terms, with the outer two being perfect squares and the middle one equal to $\pm2ab$ . If neither pattern is visible at first glance, pull out any common numerical or variable factor — the pattern is often hidden underneath one, as in $50x^2 - 8y^2 = 2(25x^2 - 4y^2)$ .

Building speed. Identity-based factorisation lives or dies on spotting perfect squares and cubes quickly by sight: $1, 4, 9, 16, 25, 36, 49, 64, 81, 100$ are the squares of $1$ through $10$ , and $1, 8, 27, 64, 125$ are the cubes of $1$ through $5$ . The same recognition applies to algebraic terms — $16x^2$ is $(4x)^2$ , and $x^4$ is $(x^2)^2$ — so training the eye on plain numbers pays off directly on algebra.

Why the reverse direction feels harder. Expanding $(a+b)^2$ is mechanical — multiply it out and collect terms. Factorising asks you to run that process in reverse, starting from the collected, expanded form and reconstructing the original brackets without being told which identity produced them. This is why the checking habit matters so much: after every factorisation, re-expand the answer and confirm it reproduces the original expression term for term. A wrong guess at the identity, or a sign slip inside it, shows up immediately once you check by expanding back.

Figure

Two branching paths showing how a factorisation problem is sorted into the difference of squares path or the perfect square trinomial path based on the number of terms and the sign pattern

Derivation AFactorising 50x squared minus 8y squared using the difference of squares

The goal is to show, in full arithmetic detail, exactly how removing a hidden common factor and then matching the difference-of-squares identity turns an expression that looks unfactorable at first glance into a clean product of two linear brackets, and to show why skipping the common-factor step causes the pattern to stay invisible.

Step 1. Scan both terms for a common factor before touching any identity. In $50x^2 - 8y^2$ , the coefficients $50$ and $8$ share a factor of $2$ , and there is no common variable factor since one term has $x^2$ and the other has $y^2$ . Pulling out the $2$ gives $50x^2 - 8y^2 = 2(25x^2 - 4y^2)$ .

Step 2. Test the bracket against the difference-of-squares pattern. $25x^2 = (5x)^2$ and $4y^2 = (2y)^2$ are both perfect squares, and they are joined by a minus sign, so the bracket fits $a^2 - b^2$ with $a = 5x$ and $b = 2y$ .

Step 3. Apply the identity $a^2 - b^2 = (a-b)(a+b)$ directly: $25x^2 - 4y^2 = (5x-2y)(5x+2y)$ .

Step 4. Reattach the common factor removed in Step 1 to get the complete factorisation: $50x^2 - 8y^2 = 2(5x-2y)(5x+2y)$ . Check: expanding back, $2(5x-2y)(5x+2y) = 2(25x^2 - 4y^2) = 50x^2 - 8y^2$ , which matches the original expression exactly.

Worked example

Factorise $9x^2 - 30xy + 25y^2$ .

Step 1. Check the outer two terms for perfect squares. $9x^2 = (3x)^2$ and $25y^2 = (5y)^2$ , so $a = 3x$ and $b = 5y$ .

Step 2. Check the middle term against $\pm 2ab$ . $2ab = 2 \times 3x \times 5y = 30xy$ . The given middle term is $-30xy$ , matching $-2ab$ , so the identity to use is $a^2 - 2ab + b^2 = (a-b)^2$ .

Step 3. Write the factorised form: $9x^2 - 30xy + 25y^2 = (3x-5y)^2$ .

Check: expanding $(3x-5y)^2 = 9x^2 - 30xy + 25y^2$ , which matches.

Final answer: $9x^2 - 30xy + 25y^2 = (3x-5y)^2$ .

Competitive stretch

Factorise $x^3 - 27$ using a cube identity. Rewrite $27 = 3^3$ , so $x^3 - 27 = x^3 - 3^3$ . With $a = x$ and $b = 3$ in $a^3 - b^3 = (a-b)(a^2+ab+b^2)$ , this gives $x^3 - 27 = (x-3)(x^2+3x+9)$ . Notice that $x^2+3x+9$ is not $(x-3)^2$ — the second bracket in a cube identity always carries a $+ab$ middle term, never $\pm2ab$ , and it generally cannot be factorised any further at Class 8 level.

Common traps

Misreading the middle-term sign. Confusing $a^2+2ab+b^2=(a+b)^2$ with $a^2-2ab+b^2=(a-b)^2$ puts the wrong sign inside the bracket. Always check whether the middle term is $+2ab$ or $-2ab$ first.
Forgetting to remove a common factor first. $50x^2-8y^2$ does not look like a clean difference of squares until the factor of $2$ is pulled out — trying the identity directly on $50x^2$ and $8y^2$ fails, since neither is itself a perfect square.
Assuming every three-term expression is a perfect-square trinomial. The middle term must equal exactly $\pm2ab$ ; if it does not, the expression is a general trinomial needing the middle-term-splitting method instead, not a forced identity.
Sign slip in the cube identities. In $a^3-b^3=(a-b)(a^2+ab+b^2)$ , students sometimes wrongly write $-ab$ in the second bracket; the correct middle sign there is always positive, for both sum and difference of cubes.

Ask the tutor: if you are unsure which identity fits a given expression, ask the AI tutor to walk through a few mixed practice problems with you.

How do I know whether to use the square identities or the cube identities?

Count the terms and look at the powers. If there are exactly two terms and each is a perfect square (like $x^2$ or $9y^2$ ) joined by a minus, it is the difference of squares. If there are three terms with the outer two as perfect squares and the middle matching $\pm2ab$ , it is a perfect-square trinomial. If instead there are two terms that are each perfect cubes (like $x^3$ or $27$ ), joined by plus or minus, reach for the sum or difference of cubes identity instead.

Why does the common-factor step matter so much here?

Many expressions are built to disguise the identity underneath a coefficient. $50x^2-8y^2$ does not look like a difference of squares at all until you divide out the shared factor of $2$ , after which $25x^2-4y^2$ is obviously $(5x)^2-(2y)^2$ . Skipping this step is the single most common reason students fail to see an identity that is actually sitting right there — always scan for a common factor before checking for a pattern.

What if the middle term in a three-term expression does not equal 2ab exactly?

Then it is not a perfect-square trinomial, and forcing the identity will give a wrong factorisation. A general trinomial like $x^2+7x+12$ needs the separate middle-term-splitting method, which is a different topic. Before applying $(a\pm b)^2$ , always verify the middle term numerically: compute $2ab$ from the square roots of the outer terms and compare it exactly against what is given.

Factorising with identities

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads