The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular subject.

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6 0

3 years ago

Factor completely <br> 4(x+1)^2/3 + 12(x+1)^-1/3

wolverine [178]

$\bf a^{\frac{{ n}}{{ m}}} \implies \sqrt[{ m}]{a^{ n}} \qquad \qquad
\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}\\\\
\left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}
\\\\
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%4(x+1)^2/3 + 12(x+1)^-1/3
4(x+1)^{\frac{2}{3}}+12(x+1)^{-\frac{1}{3}}\implies 4(x+1)^{\frac{2}{3}}+12\cfrac{1}{(x+1)^{\frac{1}{3}}}
\\\\\\$

$\bf 4(x+1)^{\frac{2}{3}}+\cfrac{12}{(x+1)^{\frac{1}{3}}}\impliedby \textit{so, our LCD is }(x+1)^{\frac{1}{3}}
\\\\\\
\cfrac{4(x+1)^{\frac{2}{3}}\cdot (x+1)^{\frac{1}{3}}+12}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4(x+1)^{\frac{2}{3}+\frac{1}{3}}+12}{(x+1)^{\frac{1}{3}}}
\\\\\\
\cfrac{4(x+1)^{\frac{3}{3}}+12}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4(x+1)+12}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4x+4+12}{(x+1)^{\frac{1}{3}}}
\\\\\\
\cfrac{4x+16}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4(x+4)}{\sqrt[3]{x+1}}$

3 0

3 years ago

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