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Double brackets: Expanding and factorising

Double brackets: Expanding and factorising

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Summary

Double brackets: Expanding and factorising

In a nutshell

Double brackets consist of two binomial expressions multiplied together, e.g. (x+1)(y3)(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.

Maths; Algebra; KS4 Year 10; Double brackets: Expanding and factorising

(x+3)(x+2)=x2First+2xOutside+3xInside+6Last=x2+5x+6\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​​
xx​​
+1+1​​
xx​​
x2x^2​​
1x1x​​
+2+2​​
2x2x​​
22​​

Add the terms and simplify.

​​​(x+1)(x+2)=x2+x+2x+2=x2+3x+2\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)(y4)(y+1)(y-4)


Multiply out then add the terms to simplify.

(y+1)(y4)=y24y+y4=y23y4\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)(3x7)(2x+5)(3x-7)


Multiply out then add the terms to simplify.

(2x+5)(3x7)=6x214x+15x35=6x2+x35\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 

ax2+bx+cax^2 + bx +c

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

Procedure

1.
Ensure the quadratic is in the form ax2+bx+cax^2 + bx + c and identify a,ba,b and cc​​
2. 
Set up the answer with two brackets and an xx at the start of each bracket, as follows (x)(x)(x \qquad)(x \qquad)​​
3.
Find two numbers which multiply to give cc and add to give bb​​
4.
Fill these numbers in the double brackets 


Example 3

Factorise 

x2+10x+24x^2 + 10x + 24


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

6×4=24 and 6+4=106 \times 4 = 24 \space and \space 6 + 4 = 10, so our numbers are 6 and 46 \space and \space 4.

x2+10x+24=(x+6)(x+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 00​, then it can be solved for xx.​


Example 4

Solve

x2+4x21=0x^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 xx will make the bracket equal to 00.​

x2+4x21=0(x+7)(x3)=0x=7,x=3\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. 

x2y2=(x+y)(xy)x^2 - y^2 = (x+y)(x-y)​​


Example 5

Factorise 9a24b29a^2-4b^2


9a29a^2 is the square of 3a3a. 4b24b^2 is the square of 2b2b. Write 3a3a as the first term in both brackets and 2b2b as the last term in both brackets. One bracket should have addition, the other should have subtraction.

9a24b2=(3a+2b)(3a2b)9a^2 - 4b^2 = \underline{(3a+2b)(3a-2b)}

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Exercises

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FAQs - Frequently Asked Questions

How do you factorise into double brackets?

How do you multiply out brackets?

What do double brackets mean?

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