# How do you simplify (4g^2 - 64g + 252)/(g-7)?

Apr 7, 2015

Let's Factorise the $\textcolor{red}{N U M E R A T O R}$ first.

The Numerator is $\textcolor{red}{4 {g}^{2} - 64 g + 252}$

$= 4 \left({g}^{2} - 16 g + 63\right)$ (4 was the factor common to all terms)

Now we need to factorise $\textcolor{b l u e}{{g}^{2} - 16 g + 63}$

We can use Splitting the Middle Term technique to factorise this.

It is in the form $a {x}^{2} + b x + c$ where $a = 1 , b = - 16 , c = 63$

To split the middle term, we need to think of two numbers ${N}_{1} \mathmr{and} {N}_{2}$ such that:
${N}_{1} \cdot {N}_{2} = a \cdot c \mathmr{and} {N}_{1} + {N}_{2} = b$
${N}_{1} \cdot {N}_{2} = \left(1\right) \cdot \left(63\right) \mathmr{and} {N}_{1} + {N}_{2} = 16$
${N}_{1} \cdot {N}_{2} = 63 \mathmr{and} {N}_{1} + {N}_{2} = - 16$

After Trial and Error, we get ${N}_{1} = - 7 \mathmr{and} {N}_{2} = - 9$
$\left(- 7\right) \cdot \left(- 9\right) = 63$ and $\left(- 7\right) + \left(- 9\right) = - 16$

So we can write the expression in blue as
$\textcolor{b l u e}{{g}^{2} - 7 g - 9 g + 63}$
$= g \left(g - 7\right) - 9 \left(g - 7\right)$
$= \left(g - 7\right) \cdot \left(g - 9\right)$

The Numerator can be written as $\textcolor{red}{4 \left(g - 7\right) \cdot \left(g - 9\right)}$

The expression we have been given is
$\frac{4 {g}^{2} - 64 g + 252}{g - 7}$

After the numerator was factorised, the Expression can now be written as :

$\frac{4 \left(g - 7\right) \cdot \left(g - 9\right)}{g - 7}$

$= \frac{4 \cdot \cancel{\left(g - 7\right)} \cdot \left(g - 9\right)}{\cancel{\left(g - 7\right)}}$

$= 4 \cdot \left(g - 9\right)$

$\frac{4 {g}^{2} - 64 g + 252}{g - 7}$ = $4 \cdot \left(g - 9\right)$