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

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Answer 1 · 2018-06-02T04:08:01

$132$ $132$

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

$6 (z^{2} + 6)$ $6(z^2+6)$

Plugging in $z = - 4$ $z=-4$ , we get

$6 ({(- 4)}^{2} + 6)$ $6((-4)^2+6)$

$\Rightarrow 6 (16 + 6)$ $=>6(16+6)$

$\Rightarrow 6 (22)$ $=>6(22)$

$\Rightarrow 132$ $=>132$

Hope this helps!

Answer 2 · 2018-06-02T04:57:55

$132$ $132$

As per the question, we have

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

$:.$ At $z = -4$ , we get

$6[(-4)^2 + 3(2)]$

$= 6(16 + 6)$

$= 6(22)$

$= 132$

Hence, the answer.

How do you simplify 6[(z^2 +3 (0+2)]6[(z2+3(0+2)]6[(z^2 +3 (0+2)] when z is -4?