SPI Mathematics · Algebraic Identities and Simplification

Factorising with Common Factors and Identities

An expression is factorised by taking out what its terms share, by grouping, or by matching it to an identity such as (a + b)², (a − b)² or a² − b².

All Algebraic Identities and Simplification notes

Why this matters

Eleven questions in the bank, eight EASY and three MODERATE. Each gives an expression and four candidate factorisations, so the quickest check is to multiply an option back out. The marks go on a dropped common factor and on a sum of squares treated as if it factorised.

Concept 1 of 4: Taking out a common factor and factorising by grouping

If every term carries the same factor, take it out once and write what is left in a bracket. With four terms that share nothing overall, pair them so that each pair shares a factor; the same bracket then appears in both pairs and comes out as well.

Definition

Common factor: $ab + ac = a(b + c)$ . Take out the largest number and the highest power of each letter that every term has.
A whole bracket can be the common factor: $x(p + q) + y(p + q) = (x + y)(p + q)$ .
Grouping: reorder the four terms so that each pair shares a factor, factorise each pair, then take out the common bracket.
If a pair starts with a minus, take the minus out with its factor so the brackets match.

Common factor

ab + ac = a(b + c)

Worked example

Factorise

2ax - 3by - 3bx + 2ay

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionEasy

Example 1 · Algebraic Identities and Simplification · Factorising with Common Factors and Identities

Factorise: a(b - c) - d(b - c)

Giving the common bracket alone

x(p + q) + y(p + q)

(x + y)(p + q)

. The shared bracket is one factor and what is left is the other; neither one alone is the factorisation.

Pairing terms that share nothing

ab + cd + ad + bc

, pair ab with ad and cd with bc:

a(b + d) + c(d + b) = (a + c)(b + d)

. Taking the terms in the printed order gives pairs with no common factor.

Drill 2 more on taking out a common factor and factorising by grouping

Concept 2 of 4: The identity for (x + a)(x + b)

Multiplying

(x + a)(x + b)

always gives the same pattern:

x^2

, then the sum of a and b times x, then their product. Read the pattern forwards to expand and backwards to factorise.

Definition

$(x + a)(x + b) = x^2 + (a + b)x + ab$ .
To factorise $x^2 + px + q$ , find a and b with $a + b = p$ and $ab = q$ .
It works with a second letter too: $x^2 + 8xy + 15y^2 = (x + 3y)(x + 5y)$ .
Two expressions equal for every x have equal coefficients, so you can find an unknown coefficient by expanding the other side.

Product of two binomials

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

Worked example

Factorise

p^2 + 10pq + 21q^2

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionEasy

Example 2 · Algebraic Identities and Simplification · Factorising with Common Factors and Identities

x^2 + kx + 6 = (x + 2)(x + 3)

, for all x, then the value of k is

Getting the middle sign wrong

(x + 4)(x - 7)

has middle term

(4 - 7)x = -3x

and constant

-28

. Add a and b with their signs, then multiply them with their signs.

Drill 2 more on the identity for (x + a)(x + b)

Concept 3 of 4: Perfect square identities

A square of a two-term bracket has three terms: the two squares and twice their product in the middle. When an expression shows two perfect squares and a middle term equal to twice the product of their roots, it is a perfect square.

Definition

$(a + b)^2 = a^2 + 2ab + b^2$ and $(a - b)^2 = a^2 - 2ab + b^2$ .
To spot one: are the first and last terms squares, and is the middle term twice the product of their roots?
Take out any common number first: $2x^2 + 4x + 2 = 2(x + 1)^2$ .
Dividing a perfect square by its root bracket leaves the bracket: $(x + 3)^2 \div (x + 3) = x + 3$ .

Perfect squares

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

Worked example

Factorise

25x^2 + 30x + 9

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionEasy

Example 3 · Algebraic Identities and Simplification · Factorising with Common Factors and Identities

Factorise:

36y^2 - 36y + 9

Dropping the common factor

50x^2 - 20x + 2 = 2(5x - 1)^2

. Leaving out the 2 gives

(5x - 1)^2

, which is only

25x^2 - 10x + 1

Forgetting the middle term

(a + b)^2

is not

a^2 + b^2

(x + 5)^2 = x^2 + 10x + 25

: the

10x

is twice the product.

Drill 1 more on perfect square identities

Concept 4 of 4: Difference of two squares

One square minus another always splits into a sum times a difference. Many expressions hide a difference of squares: take out a common number first, or add and subtract a term to make a perfect square appear.

Definition

$a^2 - b^2 = (a + b)(a - b)$ . Only a difference factorises this way; $a^2 + b^2$ does not.
Take out a common number first: $2x^2 - 18 = 2(x + 3)(x - 3)$ .
Completing the square: for an expression like $x^4 + kx^2 + c$ , add and subtract a term to build a perfect square, then use the difference of squares.
$x^2 + \frac{1}{x^2} = \left(x - \frac{1}{x}\right)^2 + 2$ , so $x^2 + \frac{1}{x^2}$ minus a number can become a difference of squares.

Difference of squares

a^2 - b^2 = (a + b)(a - b) \qquad x^2 + \frac{1}{x^2} = \left(x - \frac{1}{x}\right)^2 + 2

Worked example

Factorise

x^4 + x^2 + 1

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 4 · Algebraic Identities and Simplification · Factorising with Common Factors and Identities

The factors of

x^4 + 2x^2 + 9

is:

Factorising a sum of squares

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

, but

a^2 + b^2

has no such factors.

x^2 + 9

cannot be written as

(x + 3)(x - 3)

, which is

x^2 - 9

Calling a hidden difference 'not possible'

x^4 + 4

looks like a sum, but

x^4 + 4 = (x^2 + 2)^2 - 4x^2 = (x^2 + 2x + 2)(x^2 - 2x + 2)

. Try completing the square before ruling out factors.

Drill 2 more on difference of two squares

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

Taking out a common factor and factorising by grouping
Common factor
$ab + ac = a(b + c)$
The identity for (x + a)(x + b)
Product of two binomials
$(x + a)(x + b) = x^2 + (a + b)x + ab$
Perfect square identities
Perfect squares
$(a + b)^2 = a^2 + 2ab + b^2 \qquad (a - b)^2 = a^2 - 2ab + b^2$
Difference of two squares
Difference of squares
$a^2 - b^2 = (a + b)(a - b) \qquad x^2 + \frac{1}{x^2} = \left(x - \frac{1}{x}\right)^2 + 2$

Watch out for (7)

Giving the common bracket alone
→ Taking out a common factor and factorising by grouping
Pairing terms that share nothing
→ Taking out a common factor and factorising by grouping
Getting the middle sign wrong
→ The identity for (x + a)(x + b)
Dropping the common factor
→ Perfect square identities
Forgetting the middle term
→ Perfect square identities
Factorising a sum of squares
→ Difference of two squares
Calling a hidden difference 'not possible'
→ Difference of two squares

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