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}}

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

Answers (1)

2021-02-20
a)Given:
\(x^{\frac{5}{2}}-9x^{\frac{1}{2}}\)
On simplification, we get
\(x^{\frac{5}{2}}-9x^{\frac{1}{2}}=x^{2}\cdot x^{\frac{1}{2}}-9x^{\frac{1}{2}} \begin{bmatrix}\because x^{m} \cdot x^{n} = x^{m+n} \\\Rightarrow x^{2} \cdot x^{\frac{1}{2}}=x^{2+\frac{1}{2}}=x^{\frac{5}{2}} \end{bmatrix}\)
\(=x^{\frac{1}{2}}(x^{2}-9)\)
\(= x^{\frac{1}{2}} (x^{2} - 3^{2})\)
\(=x^{\frac{1}{2}}(x-3)(x+3)[\because a^{2}-b^{2})=(a-b)(a+b)]\)
b)Given:
\(=x^{-\frac{3}{2}}+2x(-\frac{1}{2})+x^{\frac{1}{2}}ZSL
On simplification, we get
\(x^{-\frac{3}{2}}+2x(-\frac{1}{2})+x^{\frac{1}{2}}=x^{-\frac{3}{2}}+2x(-\frac{3}{2}) \cdot x + x^{\frac{3}{2}} \cdot x^{2}[(\because x^{m} \cdot x^{n}=x^{m+n}(\Rightarrow x^{-\frac{3}{2}} \cdot x=x^{-\frac{3}{2}+1=x^{-\frac{1}{2}}})\)
And \(x^{-\frac{3}{2}} \cdot x^{2}=x^{-\frac{3}{2}+2=x^{\frac{1}{2}}}\)
\(x^{\frac{3}{2}}(1+2x+x^{2})\)
\(=x^{-\frac{3}{2}}(1+x)^{2}[\because(1+x)^{2}=1+2x+x^{2}]\)
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