# Reviewing for precalc and I dont understand how to completely factor expressions with rational exponents. a) x^frac{5}{2} - 9x^{frac{1}{2}} b) x^{-frac{2}{3}}+2x(-frac{1}{2})+x^{frac{1}{2}}

Reviewing for precalc and I dont understand how to completely factor expressions with rational exponents.
$a\right){x}^{\frac{5}{2}}-9{x}^{\frac{1}{2}}$
$b\right){x}^{-\frac{2}{3}}+2x\left(-\frac{1}{2}\right)+{x}^{\frac{1}{2}}$
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Layton

a)Given:
${x}^{\frac{5}{2}}-9{x}^{\frac{1}{2}}$
On simplification, we get
${x}^{\frac{5}{2}}-9{x}^{\frac{1}{2}}={x}^{2}\cdot {x}^{\frac{1}{2}}-9{x}^{\frac{1}{2}}\left[\begin{array}{c}\because {x}^{m}\cdot {x}^{n}={x}^{m+n}\\ ⇒{x}^{2}\cdot {x}^{\frac{1}{2}}={x}^{2+\frac{1}{2}}={x}^{\frac{5}{2}}\end{array}\right]$
$={x}^{\frac{1}{2}}\left({x}^{2}-9\right)$
$={x}^{\frac{1}{2}}\left({x}^{2}-{3}^{2}\right)$
$={x}^{\frac{1}{2}}\left(x-3\right)\left(x+3\right)\left[\because {a}^{2}-{b}^{2}\right)=\left(a-b\right)\left(a+b\right)\right]$
b)Given:
$={x}^{-\frac{3}{2}}+2x\left(-\frac{1}{2}\right)+{x}^{\frac{1}{2}}$
On simplification, we get
${x}^{-\frac{3}{2}}+2x\left(-\frac{1}{2}\right)+{x}^{\frac{1}{2}}={x}^{-\frac{3}{2}}+2x\left(-\frac{3}{2}\right)\cdot x+{x}^{\frac{3}{2}}\cdot {x}^{2}\left[\left(\because {x}^{m}\cdot {x}^{n}={x}^{m+n}\left(⇒{x}^{-\frac{3}{2}}\cdot x={x}^{-\frac{3}{2}+1={x}^{-\frac{1}{2}}}\right)$
And ${x}^{-\frac{3}{2}}\cdot {x}^{2}={x}^{-\frac{3}{2}+2={x}^{\frac{1}{2}}}$
${x}^{\frac{3}{2}}\left(1+2x+{x}^{2}\right)$
$={x}^{-\frac{3}{2}}\left(1+x{\right)}^{2}\left[\because \left(1+x{\right)}^{2}=1+2x+{x}^{2}\right]$