# Question #b9d0a

Sep 28, 2017

$2 {\left(x - 4\right)}^{2} - 8$

#### Explanation:

The first step to complete the square is to make the coefficient of ${x}^{2}$ 1. This is done by factorising the 2 out of the equation.

$2 {x}^{2} - 16 x + 24$
$2 \left({x}^{2} - 8 x + 12\right)$

Now we complete the square by halving the coefficient of $x$ and adding and subtracting this number squared (this effectively does not change the equation as in total zero is being added).

$2 \left[{x}^{2} - 8 + {\left(- 4\right)}^{2} - {\left(- 4\right)}^{2} + 12\right]$

This is the completed square before being factored:

${x}^{2} - 8 x + {\left(- 4\right)}^{2}$

Simply take the $\left(- 4\right)$ and place inside the brackets which are being squared (As you would to factorise a normal equation). This is your completed square. You also need to collect the other values in the equation.

$2 \left[{\left(x - 4\right)}^{2} - {\left(- 4\right)}^{2} + 12\right]$

Evaluate the constants

$2 \left[{\left(x - 4\right)}^{2} - 16 + 12\right]$
$2 \left[{\left(x - 4\right)}^{2} - 4\right]$

This is your answer, you can take the constant out of the [] brackets if you like.

$2 {\left(x - 4\right)}^{2} - 8$