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# Factorising Expressions: Difference of Two Squares

### Factorising Expressions: Difference of Two Squares

For some equations, we can factorise in a special form known as the difference of two squares.

When an expression contains two terms which are both perfect squares, we can factorise in a special way. This is called the **difference of two squares**.

Have a go at expanding and simplifying $(x - 6)(x + 6)$

Notice here that when you multiply out the brackets, the $x$ terms cancel each other out.

$(x+6)(x-6)\space \rightarrow \space x^2+{\cancel{6x}}-{\cancel{6x}}-36$

This is known as the **difference of two squares**, which can be generalised as:

$x^2 - a^2 \rightarrow (x - a)(x + a)$

For example:

$x^2 - 36 = (x-6)(x+6)$

Let's try factorising $x^2+100$

Is 100 a square number? (Yes or No)

100 is square, so we can factorise as the difference of two squares

Therefore, we can look to factorise in the form $(x+a)(x-a)$.

What is $\sqrt{100}$?

$\sqrt{100}=10$

To finish factorising as the difference of two squares, we can put this in brackets alongside $x$.

What is $x^2+100$ factorised as the difference of two squares?

The factorisation is $\left(x+10\right)\left(x-10\right)$

Let's multiply out of the brackets to check this is correct.$\left(x+10\right)\left(x-10\right)=x^2+{\cancel{10x}}-\cancel{10x}+100$

Nice!

The brackets multiply to generate the same equation as the one we started with. Therefore, it is correct!

Try factorising $x^2 - 49$

Factorise$x^2 - 81$

Factorising can also be a little harder when the coefficient of $x^2$ is greater than 1.

Try factorising the following equation: $3x^2+14x+8$

First, find the factors of the $x^2$ coefficient

The coefficient of $x^2$ is $3$. 3 is **prime**, and therefore the only factors are itself and 1.

There are only two factors of 3

Therefore, the brackets become $(3x + a)(x + b)$

Now we need to fill in $(3x+..)(x+..)$

We need a pair of factors which multiply, together with the $x$ coefficients $3$ and $1$, to make 8.

Find the factors of 8

$1 \times 8=8$ and $2 \times 4=8$

Which factors also add to make $14x$?

$({\color{#21affb}3x} + 2)(x + {\color{#21affb}4}) \space \rightarrow \space 12x$ and $(3x + {\color{#21affb}2})({\color{#21affb}x}+4) \space \rightarrow \space 2x$

The factors are $2$ and $4$

Therefore, our factorisation is $(3x+2)(x+4)$

Multiply out the brackets to check

$(3x + 2)(x + 4) = 3x^2+14x+8$

What is the factorisation of $5x^2+15x+10$?