Chapter A.10, Problem 102AYU

To determine

Factor the expression ${\left(x^2+4\right)}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+x.\frac{4}{3}.{\left(x^2+4\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}.2x$ and express the answer so that only positive exponents occur.

Answer to Problem 102AYU

The factorized form of the expression ${\left(x^2+4\right)}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+x.\frac{4}{3}.{\left(x^2+4\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}.2x$ is $\frac{1}{3}{\left(x^2+4\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\left(11x^2+12\right)$.

Explanation of Solution

Given:

It is asked to determine the factor of each expression and express the answer that contains only positive exponents.

Calculation:

Consider ${\left(x^2+4\right)}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+x.\frac{4}{3}.{\left(x^2+4\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}.2x$,

${\left(x^2+4\right)}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+x.\frac{4}{3}.{\left(x^2+4\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}.2x=\left(x^2+4\right){\left(x^2+4\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+x.\frac{4}{3}.{\left(x^2+4\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}.2x={\left(x^2+4\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\left[x^2+4+\frac{8x^2}{3}\right]=\frac{1}{3}{\left(x^2+4\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\left(11x^2+12\right)$