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

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Answer 1 · 2018-06-02T04:08:01

#132#

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

#6(z^2+6)#

Plugging in #z=-4#, we get

#6((-4)^2+6)#

#=>6(16+6)#

#=>6(22)#

#=>132#

Hope this helps!

Answer 2 · 2018-06-02T04:57:55

#132#

As per the question, we have

#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.