I have two expressions which I've derived by taking the derivative. Both of these expressions are from separate problems, and in both of these instances I'm having difficulty understanding the algebra that brings the expression to a factorized form. They are below:
Problem #1
Given: $$f(x)=5(2x+1)(2x-3)^4(x^2+x+1)^4+8(2x-3)^3(x^2+x+1)^5$$
Factor/Simplify to: $$f(x)=(2x-3)^3(x^2+x+1)^4(28x^2-12x-7)$$
Problem #2
Given: $$h(t) = (t+1)^{2/3}(3)(2t-1)^2(4t)+\frac{2}{3}(t+1)^{-1/3}(2t-1)^3$$
Factor/Simplify to: $$\frac{2}{3}(t+1)^{-1/3}(2t-1)^2(18t^2+20t-1)$$
If you could please help explain how to get from what I've given to the factorized form of the answer, I would really appreciate it. I know part of the process is to identify the greatest common factor (if it exists), group the like terms, and multiply by their factor. But I don't have anything close enough to the answer worthy of putting into mathjax for either of these problems. Thanks for your help.
Problem 1
$f(x)=5(2x+1)(2x-3)^4(x^2+x+1)^4+8(2x-3)^3(x^2+x+1)^5$
As in above equation we have $(2x-3)^3$ and $(x^2+x+1)^4$ common in both terms so take them out. And solve the remaining.
$f(x)=(2x-3)^3(x^2+x+1)^4\left[(5(2x+1)(2x-3)+8(x^2+x+1)\right]$
Problem 2
$h(t) = (t+1)^{2/3}(3)(2t-1)^2(4t)+\frac{2}{3}(t+1)^{-1/3}(2t-1)^3$
As in above equation we have $(t+1)^{-1}{3}$, 2 and $(2t-1)^3$ common in both terms so take them out. And solve the remaining.
$2(t+1)^{-1/3}(2t-1)^2\left[6t(t+1)+\frac{1}{3}(2t-1)\right]$