Originally Posted by

**11rdc11** I'm want to help but just not sure how to explain it. Umm lets try again.

$\displaystyle 3(2x-1)^{\frac{1}{2}}(x+1)^2-2(2x-1)^{\frac{3}{2}}(x+1)$

Ok now lets see what can be factored out of it. You can take out an (x+1)

$\displaystyle (x+1)[3(2x-1)^{\frac{1}{2}}(x + 1)-2(2x-1)^{\frac{3}{2}}]$

It can still be factored some more by factoring out $\displaystyle (2x-1)^{\frac{1}{2}}$

$\displaystyle (x+1)(2x-1)^{\frac{1}{2}}[3(x + 1)-2(2x-1)]$

which now you simply

$\displaystyle (x+1)(2x-1)^{\frac{1}{2}}[3x + 3-4x+2)]$

simplfied more

$\displaystyle (x+1)(2x-1)^{\frac{1}{2}}(-x + 5)$

Hope this helps and fixed a mistake in my original post