Question

16/35
factor completely
$r^2 - r - 56$
$4(r + 3)(r + 9)$ $3(r - 3)(r + 8)$ not factorable $(r + 7)(r - 8)$

Explanation:

Step1: Identify $a,b,c$ for $ax^2+bx+c$

For $r^2 - r - 56$, $a=1$, $b=-1$, $c=-56$

Step2: Find pair with product $ac$, sum $b$

Need two numbers: product $1\times(-56)=-56$, sum $-1$. The numbers are $7$ and $-8$ (since $7\times(-8)=-56$, $7+(-8)=-1$)

Step3: Split middle term and factor

$$\begin{align*} r^2 - r - 56&=r^2 +7r -8r -56\\ &=r(r+7)-8(r+7)\\ &=(r+7)(r-8) \end{align*}$$

Answer:

D. $(r + 7)(r - 8)$