Question 84544
I need help to fully factor this problem:
3 * (2x-1)^2 * (2) * (x+3)^1/2 + (2x-1)^3 * (1/2) * (x+3)^-1/2

You're expanding, not factoring.  When you factor, you take out the lowest powers of what each term has in common:
{{{6(2x-1)^2*(x+3)^(1/2)+(1/2)*(2x-1)^3*(x+3)^(-1/2)}}}
{{{(1/2)(2x-1)^2*(x+3)^(-1/2)(12(x+3)^(1/2-(-1/2))+(2x-1)^(3-2))}}}
{{{(1/2)(2x-1)^2(x+3)^(-1/2)(12(x+3)^(2/2)+(2x-1))}}}
{{{(1/2)(2x-1)^2(x+3)^(-1/2)(12x+36+2x-1)}}}
{{{(1/2)(2x-1)^2(x+3)^(-1/2)(14x+35)}}}
{{{(1/2)(2x-1)^2(x+3)^(-1/2)*7(2x+5)}}}
{{{highlight((7/2)(2x-1)^2(x+3)^(-1/2)(2x+5))}}}
That was a good one!!!
Happy Calculating!!!!