Intermediate Algebra: Connecting Concepts through Application

$\left( 2t^{\frac{1}{2}}-5 \right)^2$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $4t-20t^{\frac{1}{2}}+25 ,$ 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 $4(25)=100$ and the value of $b$ is $-20 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -10,-10 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 4t-10t^{\frac{1}{2}}-10t^{\frac{1}{2}}+25 .\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} \left( 4t-10t^{\frac{1}{2}} \right)- \left( 10t^{\frac{1}{2}}-25 \right) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2t^{\frac{1}{2}}\left( 2t^{\frac{1}{2}}-5 \right)- 5\left( 2t^{\frac{1}{2}}-5 \right) .\end{array} Factoring the $GCF= \left( 2t^{\frac{1}{2}}-5 \right)$ of the entire expression above results to \begin{array}{l}\require{cancel} \left( 2t^{\frac{1}{2}}-5 \right)\left(2t^{\frac{1}{2}}- 5\right) \\\\= \left( 2t^{\frac{1}{2}}-5 \right)^2 .\end{array}