Expanding Double Brackets

When we expand double brackets, we are multiplying two brackets together. This process requires multiplying each term in the first bracket by every term in the second bracket.

Let’s look at some examples.

Examples

Example 1:

Expand and simplify: \left(x+3\right) \left(x+5\right)

\left(x+3\right) \left(x+5\right) = x\left(x+5\right)+ 3\left(x+5\right)\left(x+3\right) \left(x+5\right) = x\left(x+5\right)+ 3\left(x+5\right)

= x^{2} + 5x + 3x + 15= x^{2} + 5x + 3x + 15

= x^{2} + 8x + 15= x^{2} + 8x + 15

Example 2:

Expand and multiply the following:

A) (x + 3) (x + 8)

B) (x + 3) (x - 8)

C) (2x - 1) (3x + 1)

D) (5x + 3) (2x - 5)

E) (3x - 1) (3x + 1)

F) (a + b) (a - b)

G) (a + b) (a + b)

H) (a - b) (a - b)

A) (x + 3) (x + 8) = x^{2} + 8x + 3x + 24(x + 3) (x + 8) = x^{2} + 8x + 3x + 24

= x^{2} + 11x + 24= x^{2} + 11x + 24

B) (x + 3) (x - 8) = x^{2} - 8x + 3x - 24(x + 3) (x - 8) = x^{2} - 8x + 3x - 24

= x^{2} - 5x - 24= x^{2} - 5x - 24

C) (2x - 1) (3x + 1) = 6x^{2} + 2x - 3x - 1(2x - 1) (3x + 1) = 6x^{2} + 2x - 3x - 1

= 6x^{2} - x - 1= 6x^{2} - x - 1

D) (5x + 3) (2x - 5) = 10x^{2} - 25x + 6x - 15(5x + 3) (2x - 5) = 10x^{2} - 25x + 6x - 15

= 10x^{2} - 19x - 15= 10x^{2} - 19x - 15

E) (3x - 1) (3x + 1) = 9x^{2} + 3x - 3x - 1(3x - 1) (3x + 1) = 9x^{2} + 3x - 3x - 1

= 9x^{2} - 1= 9x^{2} - 1

F) (a + b) (a - b) = a^{2} - ab + ab - b^{2}(a + b) (a - b) = a^{2} - ab + ab - b^{2}

= a^{2} - b^{2}= a^{2} - b^{2}

G) (a + b) (a + b) = a^{2} + ab + ab + b^{2}(a + b) (a + b) = a^{2} + ab + ab + b^{2}

= a^{2} + 2ab + b^{2}= a^{2} + 2ab + b^{2}

H) (a - b) (a - b) = a^{2} - ab - ab + b^{2}(a - b) (a - b) = a^{2} - ab - ab + b^{2}

= a^{2} - 2ab + b^{2}= a^{2} - 2ab + b^{2}

It’s useful to memorise the three expansions:

(a + b) (a + b) = a^{2} + 2ab + b^{2}

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

(a - b) (a - b) = a^{2} - 2ab + b^{2}

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