Calculus...more like factoring

So I'm taking Calculus and it's my freshman year, there's nothing too hard about it and I understand it just fine until I get to the algebra. It's the fucking factoring that gets me, I've started to see some shit that I didn't think you could do. For example I have to find the derivative of y=(2x-3)^5 (x^5+3)^2. I apply the product and chain rule properly and get...
y'=(2x-3)^5 * 2[(x^5+3)(5x^4)+(x^5+3)^2*5(2x-3)^4*2.
However, I have no fucking clue how that becomes
y'=30(20x-3)^4(x^5-x^4+1).
I guess my factoring skills suck complete ass. Anywhere I can find some "advanced" factoring tutorials?

You can't do that, I did it when I WAS IN ALGEBRA

Its just complicated, fucking do it yourself.

Faggot, I wasn't asking anyone to do it. I'm asking for some good factoring tutorials.

(2x-3)⁵2(x^5+3)(5x^4)+(x^5+3)²5(2x-3)⁴2 =
2(x^5+3)(2x-3)⁴{(2x-3)(5x^4) + 5(x^5+3)} =
10(x^5+3)(2x-3)⁴{(2x-3)(x^4) + x^5 + 3} =
10(x^5+3)(2x-3)⁴{2x^5 - 3x^4 + x^5 + 3} =
10(x^5+3)(2x-3)⁴{3x^5 - 3x^4 + 3} =
30(x^5+3)(2x-3)⁴(x^5 - x^4 + 1)

Your original expression is a degree 14 polynomial, and so is mine.  The "solution" you somehow got is a degree 9 polynomial, so it's obviously wrong.

FUCK FACTORING THERE'S NO WAY I'M TAKING CALCULUS.

Why is calculus easy, its just when you use the basics its gets intense.

In Maple:
f := (2*x-3)^5*2*(x^5+3)*(5*x^4)+(x^5+3)^2*5*(2*x-3)^4*2;
factor(f);
     5        5    4               4
30 (x  + 3) (x  - x  + 1) (2 x - 3)