## Intermediate Algebra (6th Edition)

$3(x-1)^2$
Factoring the $GCF=3$, then the given expression, $3x^2-6x+3$, is equivalent to $3(x^2-2x+1)$.\\ The two numbers whose product is $ac= 1(1)=1$ and whose sum is $b= -2$ are $\{ -1,-1 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $3(x^2-2x+1)$, is \begin{array}{l}\require{cancel} 3(x^2-x-x+1) \\\\= 3[(x^2-x)-(x-1)] \\\\= 3[x(x-1)-(x-1)] \\\\= 3[(x-1)(x-1)] \\\\= 3(x-1)^2 .\end{array}