Question 399238
I can tell that this is used in calculus, since it is an application of the product rule to the function {{{y = (x+3)^2(x-2)^3}}} and the expression given is equal to {{{dy/dx}}}...but you'll learn later how to find the derivative, what a derivative is, and why you don't cancel the d's in {{{dy/dx}}} :)


You can factor out {{{(x+3)(x-2)^2}}} to obtain


{{{((x+3)(x-2)^2)(2(x-2) + 3(x+3))}}}


={{{((x+3)(x-2)^2)(5x + 5)}}}