Once again, i find myself stuck in the abyss of polynomials.
this one was more confusing being that i need to multiply the first term to the trinomial,
then the second term to the on till your brains you combine terms. may i get some ?

The expression (2x+2)*(x^2+3x-2) is the same as 2x*(x^2+3x-2)+2*(x^2+3x-2)

Basically we are going to have 2x times all of (x^2+3x-2) 
And then we add on 2 times all of (x^2+3x-2) 

Let's distribute through each piece

For the first part...
2x*(x^2+3x-2) = 2x*(x^2)+2x*(3x)+2x*(-2)
2x*(x^2+3x-2) = 2x^3+6x^2-4x

And then the second part..
2*(x^2+3x-2) = 2*(x^2)+2*(3x)+2*(-2)
2*(x^2+3x-2) = 2x^2+6x-4



Putting this all together, we can say
(2x+2)*(x^2+3x-2) = 2x*(x^2+3x-2)+2*(x^2+3x-2)
(2x+2)*(x^2+3x-2) = 2x^3+6x^2-4x+2x^2+6x-4
(2x+2)*(x^2+3x-2) = 2x^3+8x^2+2x-4



The final answer is 2x^3+8x^2+2x-4