GCSELevel 6-7Cambridge iGCSE

Quadratics are expressions that, in their most general form, look like

$ax^{2}+bx+c$

where $a$, $b$ and $c$ are all numbers, called coefficients.

Some quadratics factorise into double brackets, which take the form as follows:

$ax^{2}+bx+c=(px+q)(rx+s)$

where $p$, $q$, $r$ and $s$ are all numbers.

There are different rules on how to factorise quadratics, depending on if $a=1$ or $a\neq1$

## Note: Positives and Negatives

Assuming $a>0$, you can easily determine whether each bracket will contain a $+$ or a $-$ by finding which of the three cases below your quadratic falls into.

Case 1: $b>0$ and $c>0$

In this case, both brackets will contain a $+$ sign.

Example: $x^{2}+10x+24=(x+6)(x+4)$

In this example, $b=10>0$ and $c=24>0$, so both brackets contain a $+$ sign.

Case 2: $b<0$ and $c>0$

In this case, both brackets will contain a $-$ sign.

Example: $x^{2}-6x+8=(x-4)(x-2)$

In this example, $b=-6<0$ and $c=8>0$, so both brackets contain a $-$ sign.

Case 3: $c<0$

In this case, regardless of the value of $b$, one bracket will contain a $+$ sign and the other bracket will contain a $-$ sign.

Example: $x^{2}-x-6=(x+2)(x-3)$

In this example, $c=-6<0$ so one bracket contains a $+$ sign and the other bracket contains a $-$ sign.

## Factorising Quadratics ($a=1$)

Quadratics where $a=1$ are quadratics of the form:

$x^{2}+bx+c$

The brackets will be of the form $(x+p)(x+q)$ where $p$ and $q$ add to give $b$ and multiply to give $c$. In other words:

$p+q=b$

$pq=c$

Example: Factorise $x^{2}+4x-12$

We need two numbers that add to give $4$ and multiply to give $-12$.

We know that one of the numbers must be positive and one of the numbers must be negative, because $c=-12$ so $c<0$.

$\textcolor{blue}{-2}+\textcolor{red}{6}=4$

$\textcolor{blue}{-2}\times\textcolor{red}{6}=12$

So the numbers must be $\textcolor{blue}{-2}$ and $\textcolor{red}{6}$.

Putting it all together, we have:

$x^{2}+4x-12=(x-2)(x+6)$

Level 6-7GCSECambridge iGCSE

## Factorising Quadratics ($a\neq1$)

When $a\neq1$, there will be coefficients in front of the $x$ terms inside the brackets, i.e. the factorisation will take the form:

$ax^{2}+bx+c=(px+q)(rx+s)$

From this, we know that $a=pr$ and $c=qs$, so we can generate many possible pairs for these numbers. However, the only way to find the right set is to try them all and see which gives the correct value for $b$.

Example: Factorise $3x^{2}+13x+12$

Firstly, since all terms are positive, we know all of $p$, $q$, $r$ and $s$ are going to be positive.

Next, the only pair of numbers that multiply to give $3$ are $3$ and $1$, so these must be $p$ and $r$

Next, to get $12$, we have lots of options:

$1\times12=12$

$2\times6=12$

$3\times4=12$

One of these pairs will be $q$ and $s$.

Next, start expanding every possible factorisation combination, until the correct $b$ value is found.

$(3x+12)(x+1)=3x^{2}+15x+12$ has $b=15\neq13$

$(x+12)(3x+1)=3x^{2}+37x+12$ has $b=37\neq13$

$(3x+6)(x+2)=3x^{2}+12x+12$ has $b=12\neq13$

$(x+6)(3x+2)=3x^{2}+20x+12$ has $b=20\neq13$

$(3x+4)(x+3)=3x^{2}+13x+12$ has $b=13$ is the correct factorisation.

At this point, even though we have one combination left to try, we can stop because we have found the correct set of brackets.

Hence, we can conclude that $3x^{2}+13x+12=(3x+4)(x+3)$

Level 6-7GCSECambridge iGCSE

## Example 1: Factorising Quadratics ($a=1$)

Factorise $x^{2}+3x+2$

[2 marks]

Since $b=3>0$ and $c=2>0$, both brackets will contain a $+$ sign.

We need to find two numbers that add to give $3$ and multiply to give $2$.

These numbers are $1$ and $2$, so $1$ and $2$ are the numbers in the brackets.

$x^{2}+3x+2=(x+1)(x+2)$

Level 6-7GCSECambridge iGCSE

## Example 2: Factorising Quadratics ($a\neq1$)

Factorise $10x^{2}-3x-1$

[3 marks]

Since $a\neq1$, we know that this factorisation must take the form $(px+q)(rx+s)$.

We also know that $pr=10$ and $qs=-1$.

There are multiple pairs for $p$ abnd $r$:

$10\times1=10$

$5\times2=10$

For $q$ and $s$, there is only one pair: $1$ and $-1$.

Try every combination:

$(10x+1)(x-1)=10x^{2}-9x-1$

$(10x-1)(x+1)=10x^{2}+9x-1$

$(5x+1)(2x-1)=10x^{2}-3x-1$

We can stop here because we have found the correct value for $b$.

Hence, we can conclude that:

$10x^{2}-3x-1=(5x+1)(2x-1)$

Level 6-7GCSECambridge iGCSE

We need two numbers that add to give $-2$ and multiply to give $1$.

Since $b=-2<0$ and $c=1>0$, we know that both numbers will be negative.

The numbers are $-1$ and $-1$.

Hence,

$y^{2}-2y+1=(y-1)(y-1)=(y-1)^{2}$

Gold Standard Education

We need two numbers that add to give $4$ and multiply to give $-21$.

Since $c=-21<0$, we know that one number will be positive and the other will be negative.

The possible pairs to multiply to $-21$ are:

$1$ and $-21$

$-1$ and $21$

$3$ and $-7$

$-3$ and $7$

The correct pair is $-3$ and $7$, because these add to give $4$

Hence,

$x^{2}+4x-21=(x-3)(x+7)$

Gold Standard Education

We need two numbers that add to give $52$ and multiply to give $147$.

Since $b=52>0$ and $c=147>0$, we know that both numbers will be positive.

The possible pairs to multiply to $147$ are:

$1$ and $147$

$3$ and $49$

$7$ and $21$

The correct pair is $3$ and $49$, because these add to give $52$

Hence,

$x^{2}+52x+147=(x+3)(x+49)$

Gold Standard Education

We know that this will take the form $(px+q)(rx+s)$.

We also know that, since $b=13>0$ and $c=11>0$, all of $p$, $q$, $r$ and $s$ must be positive.

$pr=2$ – the only pair that makes $2$ is $1$ and $2$.

$qs=11$ – the only pair that makes $11$ is $1$ and $11$.

This gives only two possibilites: $(2x+1)(x+11)$ and $(2x+11)(x+1)$

Multiply both out to find correct answer:

$(2x+1)(x+11)=2x^{2}+23x+11$ has $b=23\neq13$

$(2x+11)(x+1)=2x^{2}+13x+11$ has $b=13$ so must be the correct answer.

Hence,

$2x^{2}+13x+11=(2x+11)(x+1)$

Gold Standard Education

We know this will take the form $(px+q)(rx+s)$

Since $c=-221<0$, we know that $p>0$, $r>0$ and one of $q$ and $s$ is positive while the other is negative.

$pr=12$, the possible pairs for which are:

$1$ and $12$

$2$ and $6$

$3$ and $4$

$qs=-221$, the possible pairs for which are:

$1$ and $-221$

$-1$ and $221$

$13$ and $-17$

$-13$ and $17$

Now start trying combinations:

$(x+1)(12x-221)=12x^{2}-209x-221$

$(x-1)(12x+221)=12x^{2}+209x-221$

$(x+13)(12x-17)=12x^{2}+139x-221$

$(x-13)(12x+17)=12x^{2}-139x-221$

$(2x+1)(6x-221)=12x^{2}-436x-221$

$(2x-1)(6x+221)=12x^{2}+436x-221$

$(2x+13)(6x-17)=12x^{2}+42x-221$

$(2x-13)(6x+17)=12x^{2}-42x-221$

$(4x+1)(3x-221)=12x^{2}-881x-221$

$(4x-1)(3x+221)=12x^{2}+881x-221$

$(4x+13)(3x-17)=12x^{2}-29x-221$

Although there are several other combinations to try, we can stop here because we have found the correct $b$ value, so we know this must be the correct factorisation.

Hence,

$12x^{2}-29x-221=(4x+13)(3x-17)$

Gold Standard Education

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