[tex]\dfrac{3}{2}=1\dfrac{1}{2}=1+\dfrac{1}{2}\\\\(x+1)^\frac{3}{2}-x^2(x+1)^\frac{1}{2}=(x+1)^{1+\frac{1}{2}}-x^2(x+1)^\frac{1}{2}\\\\=(x+1)^1\cdot(x+1)^\frac{1}{2}-x^2(x+1)^\frac{1}{2}=(x+1)(x+1)^\frac{1}{2}-x^2(x+1)^\frac{1}{2}\\\\=(x+1)^\frac{1}{2}(x+1-x^2)\\\\Answer:\boxed{(x+1)^\frac{1}{2}(-x^2+x+1)}\\\\If\ you\ want\ factor\ completely:\\\\-x^2+x+1=0\\a=-1;\ b=1;\ c=1\\\Delta=b^2-4ac\\\Delta=1^2-4\cdot(-1)\cdot1=1+4=5 \ \textgreater \ 0\\\\x_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ x_2=\dfrac{-b+\sqrt\Delta}{2a}[/tex]
[tex]x_1=\dfrac{-1-\sqrt5}{2\cdot1}=\dfrac{-1-\sqrt5}{2}\ and\ x_2=\dfrac{-1+\sqrt5}{2}\\\\\boxed{-(x+1)^\frac{1}{2}\left(x-\frac{-1-\sqrt5}{2}\right)\left(x-\frac{-1+\sqrt5}{2}\right)}[/tex]
