Multiplying Out Brackets

Section 1 of 3

Single Brackets

A bracket is a holding pen: whatever sits in front multiplies every term inside, not just the first.

Single bracket

$a(b + c) = ab + ac$.

Worked example — a number outside

Multiply out: $3(2x)$, $3(x + 2)$, $5(4x + 7y)$, $7(2a + 3b)$.

$3(2x) = 6x$

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

$5(4x + 7y) = 20x + 35y$

$7(2a + 3b) = 14a + 21b$

$6x;\ 3x+6;\ 20x+35y;\ 14a+21b$

Worked example — a letter outside

Multiply out $x(x + 3)$ and $x(x + 5) + 2(x + 5)$.

$x(x + 3) = x^2 + 3x$ (since $x \cdot x = x^2$)

$x(x+5) + 2(x+5) = x^2 + 5x + 2x + 10 = x^2 + 7x + 10$

$x^2 + 3x$ and $x^2 + 7x + 10$

A minus outside

A minus in front of a bracket is $\times(-1)$ — every sign inside flips. Most marks are lost from a dropped sign.

You try

Multiply out $-(x - 3)$, $-2(x + 5)$ and $-3(2x - 7)$.

The minus attaches to whatever comes after it — flip every sign inside.

$-(x - 3) = -x + 3$

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

$-3(2x - 7) = -6x + 21$

$-x+3;\ -2x-10;\ -6x+21$

Section 2 of 3

Double Brackets

Two brackets

$(a + b)(c + d) = ac + ad + bc + bd$. Split the first bracket, multiply each term by the whole second bracket, then collect like terms.

Worked example — split and distribute

Multiply out and simplify $(x + 3)(x + 9)$.

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

$= x^2 + 9x + 3x + 27$

$x^2 + 12x + 27$

Worked example — with negatives

Multiply out and simplify $(x - 9)(x + 6)$.

$= x(x + 6) - 9(x + 6)$

$= x^2 + 6x - 9x - 54$

$x^2 - 3x - 54$

You try

Multiply out and simplify $(2x - 3)(7x - 8)$.

Same method, just multiply more carefully. $2x(7x-8) - 3(7x-8)$.

$= 2x(7x - 8) - 3(7x - 8)$

$= 14x^2 - 16x - 21x + 24$

$14x^2 - 37x + 24$

Section 3 of 3

Squared Brackets

A square

$(a + b)^2 = (a + b)(a + b)$ — then multiply out as normal. No shortcut: $5^2 = 25$, not $5 + 5$.

Worked example — square of a sum

Multiply out and simplify $(x + 7)^2$.

$= (x + 7)(x + 7) = x(x + 7) + 7(x + 7)$

$= x^2 + 7x + 7x + 49$

$x^2 + 14x + 49$

You try

Multiply out and simplify $(2x - 3)^2$.

Write it as $(2x-3)(2x-3)$ first. Remember $(2x-3)^2$ is not $4x^2 - 9$ — the middle term matters.

$= (2x - 3)(2x - 3) = 2x(2x - 3) - 3(2x - 3)$

$= 4x^2 - 6x - 6x + 9$

$4x^2 - 12x + 9$

That’s Class 2.

Single brackets, double brackets, and squares — multiply everything out, then collect like terms. Class 3: factorising, the same road backwards.