## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$(x-3)(7x-16)$
First, since the polynomial $x^2 - 6x + 9$ is a perfect square, we can factor it using the formula $a^2-2ab+b^2=(a-b)^2$, where $a=x$ and $b=3$: $$7(x^2 - 6x + 9) + 5(x - 3)=7(x-3)^2 +5(x -3) .$$ Then, factoring $x-3$ out gives $$7(x-3)^2 +5(x -3)= (x-3)[7(x-3)+5] .$$ Finally, we distribute $7$ then combine like terms to obtain: \begin{align*} (x-3)[7(x-3)+5]&= (x-3)[7x-21+5] \\ &=(x-3)(7x-16). \end{align*}