Answer
$(2r^{\frac{1}{2}}-5)(3r^{\frac{1}{2}}+4)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
6r+7r^{\frac{1}{2}}-20
,$ find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping.
$\bf{\text{Solution Details:}}$
Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $
6(-20)=-120
$ and the value of $b$ is $
7
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
-15,8
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
6r-15r^{\frac{1}{2}}+8r^{\frac{1}{2}}-20
.\end{array}
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
(6r-15r^{\frac{1}{2}})+(8r^{\frac{1}{2}}-20)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
3r^{\frac{1}{2}}(2r^{\frac{1}{2}}-5)+4(2r^{\frac{1}{2}}-5)
.\end{array}
Factoring the $GCF=
(2r^{\frac{1}{2}}-5)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(2r^{\frac{1}{2}}-5)(3r^{\frac{1}{2}}+4)
.\end{array}
