Factorisation

What is a factor?

If a number or algebraic expression can be written as the product of smaller numbers or expressions, those smaller quantities are called its factors.

For example, 21 can be written as 21 = 1×3×7.

Therefore, 1, 3 and 7 are factors of 21.

📚 Remember: if a number is divided by one of its factors, the remainder is zero.

For example:

21 ÷ 3 = 7 → it is completely divisibledivisible, so the remainder is zero.

21 ÷ 6 → it is not completely divisible, so a remainder is left.

Therefore, 6 is not a factor of 21.

Factors of algebraic expressions

We can write ab as a×b. Therefore, both a and b are factors of ab.

Factors can also be found by using square and cube identities.

🔸 x² − y² can be written as (x + y)(x − y). Hence, (x + y) and (x − y) are factors.

🔸 x³ − y³ can be written as (x − y)(x² + xy + y²).

🔸 x³ + y³ can be written as (x + y)(x² − xy + y²).

Factorisation

In algebra, factorisation means expressing a complex algebraic expression as a product of two or more simpler expressions.

It is essentially the reverse of multiplication. If the factors are multiplied together, the original expression is obtained again.

Common methods of factorisation include:

Taking a Common Factor

If all the terms contain a common factor, take that factor outside the brackets.

🔹 ax + ay

= a(x + y)

🔹 x² + xy + 7x + 7y

= x(x + y) + 7(x + y)

= (x + y)(x + 7)

🔹 15pq + 15 + 9q + 25p

= 15pq + 25p + 9q + 15

= 5p(3q + 5) + 3(3q + 5)

= (3q + 5)(5p + 3)

👉 Click for more solved problems

👇 Practice: Factorise

🔸 xy + y + 3x + 3

🔸 pq − q + 2p − 2

🔸 6xy + 3y + 4x + 2

🔸 10xy + 2y + 5x + 1

Factorisation using algebraic identities

👉 Factorisation based on

🔹 x² − 4y²

= x² − (2y)²

= (x + 2y)(x − 2y)

🔹

=

=

=

=

🔹

=

=

=

=

👇 Practice: Factorise

🔸

🔸

🔸

🔸

👉 Factorisation based on

🔹

=

=

=

🔹 63m³ + 6m² − 12m + 8

= 64m³ − m³ + 6m² − 12m + 8

= (4m)³ − {m³ − 3.m².2 + 2.m.2² − 2³}

= (4m)³ − (m − 2)³

= {4m − (m − 2)} {(4m)² + 4m(m − 2)+(m − 2)²}

= (4m − m + 2) (16m² + 4m² − 8m + m² − 2.m.2 + 2²)

= (3m + 2)(21m² − 8m − 4m + 4)

= (3m + 2)(21m² − 12m + 4)

🔹 a³ − 9b³ + (a + b)³

= a³ − b³ + (a + b)³ − 8b³

= a³ − b³ + (a + b)³ − (2b)³

= (a − b) (a² + ab + b²) + {(a + b) − 2b}{(a + b)² + (a + b).2b + (2b)²}

= (a − b)(a² + ab + b²) + (a + b − 2b)(a² + 2ab + b² + 2ab + 2b² + 4b²)

= (a − b)(a² + ab + b²) + (a − b)(a² + 4ab + 7b²)

= (a − b){(a² + ab + b²)+(a² + 4ab + 7b²)}

= (a − b)(a² + ab + b² + a² + 4ab + 7b²)

= (a − b)(2a² + 5ab + 8b²)

🔹 32x⁴ − 500x

= 4x(8x³ − 125)

= 4x{(2x)³ − (5)³}

= 4x(2x − 5){(2x)² + (2x).5 + 5²}

= 4x(2x − 5)(4x² + 10x + 25)

👇 Practice: Factorise

🔸 t⁹ − 512

🔸 a³ + 3a²b + 3ab² + b³ − 8

🔸 x³ − 6x² + 12x − 35

🔸 AX³ − Ax³ + AX²h − Ax²h

👉 Factorisation based on x³ + y³ = (x + y)(x² − xy + y²)

🔹 27a³ + b³

= (3a)³ + (b)³

= (3a + b) {(3a)² − (3a)b + (b)²}

= (3a + b)(9a² − 3ab + b²)

👉 Middle Term Factorisation

To factorise an expression of the form ax² + bx + c:

First find the product of the coefficient a of the first term and the constant term c.

Then split ac into two numbers whose sum is b.

Finally, split the middle term and factorise by grouping.

🔹 3x² + 14x + 8

The product of 3 and 8 is 24. We can write 24 as 4×6, 3×8, or 2×12. Since 12 + 2 = 14, split the middle term 14x into 12x + 2x.

= 3x² + (12 + 2)x + 8

= 3x² + 12x + 2x + 8

= 3x(x + 4) + 2(x + 4)

= (x + 4)(3x + 2)

🔹 6x² − x − 15

Here 6 × (−15) = −90. We need two numbers whose product is −90 and whose sum is −1. The numbers are −10 and 9.

= 6x² − 10x + 9x −15

= 2x(3x − 5) + 3(3x − 5)

= (3x − 5)(2x + 3)

🔹 x² − x − 6

Here 1 × (−6) = −6. We need two numbers whose product is −6 and whose sum is −1. The numbers are −3 and 2.

= x² − 3x + 2x − 6

= x(x − 3) + 2(x − 3)

= (x − 3)(x + 2)

🔹 x² − 2ax + (a + b)(a − b)

The middle term −2ax can be written as −(a+b)x − (a-b)x, because (a+b)+(a-b)=2a.

= x² − {(a + b) + (a − b)}x + (a + b)(a − b)

= x² − (a+b)x − (a − b)x + (a + b)(a − b)

= x{x − (a + b)} − (a − b){x − (a + b)}

= {x − (a + b)}{x − (a − b)}

= (x − a − b)(x − a + b)

👇 Practice: Factorise

🔸 (a + b)² − 5a − 5b + 6

🔸 (p² − 3q²)² − 16(p² − 3q²) − 63

🔸 x⁴ + 4x² − 5

🔸 x² − bx − (a + 3b)(a + 2b)

👉 Vanishing or Trial Method

For a cubic polynomial, let the expression be f(x) and find a value of x for which f(x)=0.

If f(m)=0, then (x−m) is a factor of the polynomial.

Example: x³ − 7x − 6

Let f(x) = x³ − 7x − 6. Substituting x = −1:

Therefore, x = −1 is a zero and (x + 1) is a factor.

🔹 x³ − 7x − 6

= x³ + x² − x² − x − 6x − 6

= x²(x + 1) − x(x + 1) − 6(x + 1)

= (x + 1)(x² − x − 6)

= (x + 1){x² − 3x + 2x − 6}

= (x + 1){x(x − 3) + 2(x − 3)}

= (x + 1)(x − 3)(x + 2)

👉 Click for more solved problems