Chapter A.10, Problem 107AYU

To determine

Factor the expression $4{\left(3x+5\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\left(2x+3\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+3{\left(3x+5\right)}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\left(2x+3\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ and express the answer so that only positive exponents occur.

Answer to Problem 107AYU

The factorized form of the expression $4{\left(3x+5\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\left(2x+3\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+3{\left(3x+5\right)}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\left(2x+3\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ is ${\left(3x+5\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\left(2x+3\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(17x+27\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 $4{\left(3x+5\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\left(2x+3\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+3{\left(3x+5\right)}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\left(2x+3\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$,

$=4{\left(3x+5\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\left(2x+3\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+3{\left(3x+5\right)}^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\left(2x+3\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$={\left(3x+5\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\left(2x+3\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[4\left(2x+3\right)+3\left(3x+5\right)\right]$

$={\left(3x+5\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\left(2x+3\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(17x+27\right)$