## Intermediate Algebra (12th Edition)

$9(k+1)^2$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $(3k+5)^2-4(3k+5)+4 ,$ use substitution. Then use the factoring of trinomials in the form $x^2+bx+c.$ $\bf{\text{Solution Details:}}$ Let $z=(3k+5).$ Then the given expression is equivalent to \begin{array}{l}\require{cancel} z^2-4z+4 .\end{array} In the trinomial expression above, $b= -4 ,\text{ and } c= 4 .$ Using the factoring of trinomials in the form $x^2+bx+c,$ the two numbers whose product is $c$ and whose sum is $b$ are $\left\{ -2,-2 \right\}.$ Using these two numbers, the factored form of the expression above is \begin{array}{l}\require{cancel} (z-2)(z-2) \\\\= (z-2)^2 .\end{array} Since $z=(3k+5),$ by back substitution, the expression above is equivalent to \begin{array}{l}\require{cancel} (3k+5-2)^2 \\\\= (3k+3)^2 \\\\= [3(k+1)]^2 \\\\= 3^2(k+1)^2 \\\\= 9(k+1)^2 .\end{array}