Answer
$7(2k-1)(k-1)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
14k^2-21k+7
,$ factor first the $GCF.$ Then 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.
$\bf{\text{Solution Details:}}$
Factoring the $GCF=
7
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
7(2k^2-3k+1)
.\end{array}
Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $
2(1)=2
$ and the value of $b$ is $
-3
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
-1,-2
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
7(2k^2-1k-2k+1)
.\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}
7[(2k^2-1k)-(2k-1)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
7[k(2k-1)-(2k-1)]
.\end{array}
Factoring the $GCF=
(2k-1)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
7[(2k-1)(k-1)]
\\\\=
7(2k-1)(k-1)
.\end{array}
