# How do you simplify 6[(z^2 +3 (0+2)] when z is -4?

Jun 2, 2018

$132$

#### Explanation:

Let's simplify the inside first. $0 + 2$ obviously simplifies to $2$, and multiplying it by $3$ will yield $6$. Now, we have

$6 \left({z}^{2} + 6\right)$

Plugging in $z = - 4$, we get

$6 \left({\left(- 4\right)}^{2} + 6\right)$

$\implies 6 \left(16 + 6\right)$

$\implies 6 \left(22\right)$

$\implies 132$

Hope this helps!

Jun 2, 2018

$132$

#### Explanation:

As per the question, we have

$6 \left[{z}^{2} + 3 \left(0 + 2\right)\right]$

$\therefore$ At $z = - 4$, we get

$6 \left[{\left(- 4\right)}^{2} + 3 \left(2\right)\right]$

$= 6 \left(16 + 6\right)$

$= 6 \left(22\right)$

$= 132$