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

$(x+5)(3x+11)$
First, since the polynomial $x^2 + 10x + 25$ is a perfect squrae, we can factor is using the formula $a^2+2ab+b^2=(a+b)^2$ where $a=x$ and $b=5$ to obtain: $$3(x^2 + 10x + 25) - 4(x + 5)=3(x+5)^2 - 4(x + 5) .$$ Then, factoring $x+5$ out gives $$3(x+5)^2 - 4(x + 5)= (x+5)[3(x+5)-4]$$ DIstribute $3$ then combine like terms to obtain: \begin{align*} (x+5)[3(x+5)]-4&=(x+5)(3x+15-4)\\ &=(x+5)(3x+11) \end{align*}