2 factoring problems?

Question

Problem 1:

2x² + 32
I know that when there are three terms the constant on the right must be factored and equal, in this case, 32. Would I put 2x + x + 32 to get rid of the ² and to create a middle term to do it the way I'm familiar with?

Problem 2:

(x+1)(x-5)(x+4)
How would I go about solving this? I'm not sure how to FOIL trinomials if that's even what you need to do here...

Answer 1

1- 2x^2+32=2(x^2+16)
2- (x+1)(x-5)(x+4), first (x+1)(x-5)=
x^2-4x-5.
Then mult by (x+4),

(x^2-4x-5)(x+4)=x^3-4x^2-5x+4x^2-16x-20
=x^3-21x-20

Answer 2

2x² + 32

2(x² + 16)

This expression can be factored further with imaginary numbers in this way.

2(x + 4i)(x - 4i)

where i = √(-1)
.