Factorisation

Lesson

We have already learnt how to expand the product of two binomials using the FOIL technique, the result of which is called a **quadratic**, which is any algebraic expression where only one type of variable exists but the highest power of that variable is $2$2. For example, $2x^2-5x+1$2`x`2−5`x`+1 is a quadratic, and so is $-x^2-47$−`x`2−47. Now we're going to learn how to reverse that expansion process and factorise to get back to our original binomials. We're going to focus on a special type of quadratic called a **monic quadratic**, which just means that the coefficient for the $x^2$`x`2 part is always going to be $1$1!

Now we know that $\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab$(`x`+`a`)(`x`+`b`)=`x`2+(`a`+`b`)`x`+`a``b`

So to factorise a monic quadratic expression like $x^2+Px+Q$`x`2+`P``x`+`Q`,

1. Find two numbers, $a$`a` and $b$`b`, that have a **product** of $Q$`Q` and a **sum** of $P$`P`

- It's usually a lot easier to look at $Q$
`Q`and start listing out its factors first rather than first finding the numbers that add to make $P$`P`!

2. The factorised form should be $\left(x+a\right)\left(x+b\right)$(`x`+`a`)(`x`+`b`)

Watch Out!

- If $Q$
`Q`is positive and $P$`P`is positive, then $a$`a`and $b$`b`are positive - If $Q$
`Q`is positive and $P$`P`is negative, then $a$`a`and $b$`b`are negative - If $Q$
`Q`is negative, then one of the two numbers is positive and the other is negative, regardless of $P$`P`

Another way to factorise quadratics is called **the cross method**. This method can be used even for non-monic quadratics, so let's have a look at how it works! Let's take the example $x^2+7x+6$`x`2+7`x`+6. We know the answer must be of the form $\left(x+a\right)\left(x+b\right)$(`x`+`a`)(`x`+`b`), as only this format will give us the first term $x^2$`x`2, since the only factor pair of $x^2$`x`2 is $x$`x` & $x$`x` (Technically $x^2$`x`2 & $1$1 is another pair, but it would not help us factorise!). Then our job is just to find what $a$`a` and $b$`b` are. We know they must multiply together to give us $6$6, so let's take a look at the factor pairs of $6$6, which are $2$2 & $3$3 or $1$1 & $6$6.

The cross method works by trial and error, and we start by drawing a cross, with a possible factor pair of the first $x^2$`x`2 term on the left, and a possible factor pair of $6$6 on the right. Let's try the pair $2$2 & $3$3 for now:

Then we follow along each arrow to make a product of the numbers linked by them. For example the blue arrow gives us $3x$3`x` while the green one gives us $2x$2`x`. Together they have a sum of $5x$5`x`, but our goal is to reach $7x$7`x`, the middle term of the quadratic. That means $2$2 & $3$3 must not work. Let us try $6$6 & $1$1 then:

Hooray! This configuration does indeed give us the sum we're looking for, which is $7x$7`x`. To read off the correct factorised form of this quadratic then, simply put each circled pair in brackets and multiply them together. For example here we can factorise $x^2+7x+6$`x`2+7`x`+6 to become $\left(x+1\right)\left(x+6\right)$(`x`+1)(`x`+6).

Factorise $x^2+15x+56$`x`2+15`x`+56

**Think** about whether $a$`a` and $b$`b` are positive or negative here

**Do**

$a+b=15$`a`+`b`=15 and $a\times b=56$`a`×`b`=56

$a$`a` and $b$`b` must both then be positive

If we look at the factors of $56$56, we get these pairs:

$1$1, $56$56

$2$2, $28$28

$4$4, $14$14

$7$7, $8$8

The only pair with a sum of $15$15 is the last pair, so $a$`a` and $b$`b` must equal $7$7 and $8$8.

So $x^2+15x+56=\left(x+7\right)\left(x+8\right)$`x`2+15`x`+56=(`x`+7)(`x`+8)

Use the cross method to solve $x^2+2x-3$`x`2+2`x`−3.

**Think **about how there's two different factor pairs for $-3$−3 as it is a negative number

Do

$-3$−3 has the factor pairs $1$1 & $-3$−3 and $-1$−1 & $3$3. Let us try the first pair first:

This does not give us the answer we want which is $2x$2`x`. Let's then try the other pair:

This gives us the correct sum so the factorised form must be $\left(x-1\right)\left(x+3\right)$(`x`−1)(`x`+3).

Factorise $x^2-2x-8$`x`2−2`x`−8.

Factorise $44-15x+x^2$44−15`x`+`x`2.

Factorise the expression completely by first taking out a common factor:

$3x^2-21x-54$3`x`2−21`x`−54

Form and solve linear equations and inequations, quadratic and simple exponential equations, and simultaneous equations with two unknowns

Apply algebraic procedures in solving problems