Factorisation — Class 8

Factorisation is the reverse of multiplication. When we multiply, we combine factors to get a product. When we factorise, we break a product back into its factors. The skill matters for everything that comes after — solving equations in Class 9, simplifying algebraic fractions in Class 10, and rapidly handling polynomials in Class 11.

Think of factorisation as “unmultiplying”. $6 = 2 \times 3$ — we’ve factorised 6 into 2 and 3. $x^2 + 5x + 6 = (x+2)(x+3)$ — we’ve factorised the algebraic expression on the left into the two binomials on the right.

What is a Factor?

A factor of a number (or expression) is something that divides it exactly, without leaving a remainder.

Factors of 12: 1, 2, 3, 4, 6, 12
Factors of $6x$ : 1, 2, 3, 6, $x$ , $2x$ , $3x$ , $6x$
Factors of $x(x+1)$ : 1, $x$ , $(x+1)$ , $x(x+1)$

When we factorise an expression, we write it as a product of its factors.

Method 1: Common Factor Method

Look for a factor that appears in every term. Pull it out.

Example 1: Factorise $6x + 9$ .

Both 6 and 9 are divisible by 3. So:

$6x + 9 = 3(2x) + 3(3) = 3(2x + 3)$

Example 2: Factorise $4a^2b - 6ab$ .

Common factors: 2, $a$ , $b$ — so $2ab$ is the highest common factor (HCF).

$4a^2b - 6ab = 2ab(2a - 3)$

Always look for the largest common factor. Pulling out 2 from $4a^2b - 6ab$ gives $2(2a^2b - 3ab)$ , which is correct but not fully factorised.

Method 2: Grouping

When there’s no single common factor for all terms, group terms cleverly.

Example: Factorise $ax + ay + bx + by$ .

Group: $(ax + ay) + (bx + by) = a(x + y) + b(x + y)$

Now $(x + y)$ is the common factor: $(x + y)(a + b)$ .

Example: Factorise $x^2 + 3x + 2x + 6$ .

Group: $(x^2 + 3x) + (2x + 6) = x(x + 3) + 2(x + 3) = (x + 3)(x + 2)$ .

Method 3: Identities (Algebraic Formulas)

Memorise these — they’re the workhorses of Class 8 to Class 12.

$(a + b)^2 = a^2 + 2ab + b^2$

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

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

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

$(x + a)(x + b) = x^2 + (a + b)x + ab$

Example using $a^2 - b^2$ : Factorise $25x^2 - 16y^2$ .

Recognise $25x^2 = (5x)^2$ and $16y^2 = (4y)^2$ . So:

$25x^2 - 16y^2 = (5x + 4y)(5x - 4y)$ .

Example using $(a+b)^2$ : Factorise $x^2 + 6x + 9$ .

Notice $x^2 + 6x + 9 = x^2 + 2 \cdot 3 \cdot x + 3^2 = (x + 3)^2$ .

Example using $(x+a)(x+b)$ : Factorise $x^2 + 7x + 12$ .

Find two numbers that multiply to 12 and add to 7: 3 and 4. So:

$x^2 + 7x + 12 = (x + 3)(x + 4)$ .

Method 4: Splitting the Middle Term

For quadratic expressions $ax^2 + bx + c$ , find two numbers $p$ and $q$ such that:

$p + q = b$
$p \times q = ac$

Then split $bx$ into $px + qx$ and group.

Example: Factorise $2x^2 + 7x + 6$ .

Here $a = 2$ , $b = 7$ , $c = 6$ , so $ac = 12$ . Find two numbers that multiply to 12 and add to 7: 3 and 4.

$2x^2 + 7x + 6 = 2x^2 + 3x + 4x + 6 = x(2x + 3) + 2(2x + 3) = (2x + 3)(x + 2)$ .

Worked Examples — Easy to Hard

Easy

Factorise $5x + 10$ .

$5x + 10 = 5(x + 2)$ .

Medium

Factorise $x^2 - 9$ .

$x^2 - 9 = x^2 - 3^2 = (x + 3)(x - 3)$ .

Hard

Factorise $x^4 - 16$ .

First apply $a^2 - b^2$ : $x^4 - 16 = (x^2)^2 - 4^2 = (x^2 + 4)(x^2 - 4)$ .

Now further factorise $x^2 - 4 = (x + 2)(x - 2)$ .

So $x^4 - 16 = (x^2 + 4)(x + 2)(x - 2)$ .

(Note: $x^2 + 4$ doesn’t factor over real numbers.)

Common Mistakes to Avoid

Mistake 1: Pulling out the wrong common factor.

In $4x^2 + 6x$ , the HCF is $2x$ , not $2$ or $x$ alone. Always factorise as much as possible.

Mistake 2: Forgetting to check the sign.

$x^2 - 5x + 6 = (x - 2)(x - 3)$ . Both factors are negative because the middle term is negative AND the constant is positive. Always verify by expanding.

Mistake 3: Confusing $a^2 - b^2$ with $a^2 + b^2$ .

$a^2 - b^2$ factors as $(a + b)(a - b)$ . $a^2 + b^2$ does NOT factor over real numbers. Don’t try to use $(a + b)^2$ or anything similar.

Mistake 4: Splitting middle term wrongly.

For $x^2 + 5x + 6$ , find numbers multiplying to 6 and adding to 5 (not subtracting!). The numbers are 2 and 3, both positive.

Mistake 5: Not factorising completely.

$2x^2 - 8 = 2(x^2 - 4)$ is partially factored. Continue: $2(x^2 - 4) = 2(x + 2)(x - 2)$ . Always check if remaining factors can be factored further.

Practice Questions

Factorise $7a + 14b$ .
Factorise $x^2 + 8x + 15$ .
Factorise $9 - x^2$ .
Factorise $4x^2 - 12x + 9$ .
Factorise $ab + bc + ad + cd$ (use grouping).
Factorise $2x^2 - 7x + 6$ (split middle term).
Factorise $x^2 - 6xy + 9y^2$ (use identity).
Factorise $a^4 - b^4$ .

Q1: $7(a + 2b)$

Q2: Numbers multiplying to 15, adding to 8: 3 and 5. So $(x + 3)(x + 5)$ .

Q3: $9 - x^2 = (3 + x)(3 - x)$

Q4: $4x^2 - 12x + 9 = (2x - 3)^2$ (perfect square trinomial)

Q5: $b(a + c) + d(a + c) = (a + c)(b + d)$

Q8: $a^4 - b^4 = (a^2)^2 - (b^2)^2 = (a^2 + b^2)(a^2 - b^2) = (a^2 + b^2)(a + b)(a - b)$

FAQs

Q: How do I check if my factorisation is correct?

Multiply the factors back. If you get the original expression, you’re correct. For example, factorising $x^2 + 5x + 6 = (x + 2)(x + 3)$ . Check: $(x + 2)(x + 3) = x^2 + 5x + 6$ . ✓

Q: Are there expressions that cannot be factorised?

Yes — over real numbers, $x^2 + 1$ has no real factors. (It factors over complex numbers as $(x + i)(x - i)$ , but that’s beyond Class 8.) Class 8 problems are always designed to factorise cleanly.

Q: When should I use grouping vs splitting the middle term?

Grouping: when you have 4 or more terms with no single common factor. Splitting middle term: specifically for quadratic trinomials (3 terms, with the middle term being $bx$ ).

Q: Why is factorisation so important?

Factorisation lets you solve equations easily. If $(x + 2)(x - 3) = 0$ , then $x = -2$ or $x = 3$ . This is the foundation of solving quadratic and higher-degree equations in Class 9 onwards.

Q: What’s the difference between factorising and expanding?

Expanding goes from factors to product: $(x + 2)(x + 3) \to x^2 + 5x + 6$ . Factorising goes the other way: $x^2 + 5x + 6 \to (x + 2)(x + 3)$ . Both skills are essential.