# Double brackets: Expanding and factorising

## In a nutshell

Double brackets consist of two binomial expressions multiplied together, e.g. $(x+1)(y-3)$. You should be able to multiply out a set of double brackets as well as factorise back into double brackets.

## Multiplying out double brackets

When multiplying out double brackets, make sure to multiply every term in the first bracket with every term in the second bracket, you can use F.O.I.L., or a multiplication grid.

#### FOIL

First | Multiply the first terms of each bracket. |

Outside | Multiply the outside terms. |

Inside | Multiply the two inside terms. |

Last | Multiply the last terms of each bracket. |

*Note:** **Once the brackets have been multiplied out, simplify the terms where possible.*

$\begin{aligned}(x+3)(x+2)&= \underbrace{x^2}_{\text{First}}+\underbrace{2x}_{\text{Outside}}+\underbrace{3x}_{\text{Inside}}+\underbrace{6}_{\text{Last}} \\&=x^2+5x+6\end{aligned}$

#### Multiplication grid

$\times$ | $x$ | $+1$ |

$x$ | $x^2$ | $1x$ |

$+2$ | $2x$ | $2$ |

Add the terms and simplify.

$\begin {aligned} (x+1)(x+2) &= x^2 + x + 2x + 2 \\ &= \underline{x^2 + 3x + 2} \end {aligned}$

Example 1

*Multiply out and simplify*

$(y+1)(y-4)$

*Multiply out then add the terms to simplify.*

##### $\begin {aligned} (y+1)(y-4) &=y^2 -4y+y -4 \\ &=\underline{y^2 - 3y -4} \end {aligned}$

##### Example 2

*Multiply out and simplify*

$(2x+5)(3x-7)$

*Multiply out then add the terms to simplify.*

$\begin {aligned} (2x+5)(3x-7) &= 6x^2 - 14x + 15x - 35 \\&= \underline{6x^2 + x - 35}\end {aligned}$

## Factorising double brackets

Factorising a quadratic means putting it into double brackets. A quadratic is in the form

$ax^2 + bx +c$

When $a=1$ we can factorise a quadratic as follows.

#### Procedure

1. | Ensure the quadratic is in the form $ax^2 + bx + c$ and identify $a,b$ and $c$ |

2. | Set up the answer with two brackets and an $x$ at the start of each bracket, as follows $(x \qquad)(x \qquad)$ |

3. | Find two numbers which multiply to give $c$ and add to give $b$ |

4. | Fill these numbers in the double brackets |

##### Example 3

*Factorise *

$x^2 + 10x + 24$

*The quadratic is in the correct form where* $b=10$ *and* $c=24$.

$6 \times 4 = 24 \space and \space 6 + 4 = 10$, *so our numbers are* $6 \space and \space 4$.

$x^2 + 10x+24 = \underline{(x+6)(x+4)}$

#### Solving a quadratic

If the quadratic is in an equation, rearrange it so it is equal to $0$, then it can be solved for $x$.

##### Example 4

*Solve*

$x^2+4x-21 = 0$

*First factorise the quadratic, then the solution is the number in the bracket with the sign reversed. This is because this value of* $x$ *will make the bracket equal to* $0$.

$\begin {aligned} x^2 +4x - 21 &= 0 \\ (x+7)(x-3) &=0 \\ \underline{x = -7, x = 3}\end {aligned}$

#### Difference of two squares

The difference of two squares is a where you have one term squared minus another term squared. The expression can be factorised into double brackets.

$x^2 - y^2 = (x+y)(x-y)$

##### Example 5

*Factorise* $9a^2-4b^2$

$9a^2$ *is the square of* $3a$. $4b^2$ *is the square of* $2b$. *Write* $3a$ *as the first term in both brackets and* $2b$ *as the last term in both brackets. One bracket should have addition, the other should have subtraction.*

$9a^2 - 4b^2 = \underline{(3a+2b)(3a-2b)}$