Answer:
[tex]\huge\boxed{r^2+4r-32=(r+8)(r-4)}[/tex]
Step-by-step explanation:
METHOD 1.
[tex]r^2+4r-32\\\\\left\begin{array}{ccc}a\cdot b=-32\\a+b=4\end{array}\right\}a=8;\ b=-4\\\\r^2+8r-4r-32\qquad|\text{distribute}\\\\=r(r+8)-4(r+8)\\\\=(r+8)(r-4)[/tex]
METHOD 2.
Use the quadratic formula.
[tex]ax^2+bx+c=a(x-x_1)(x-x_2)\\\\x_1=\dfrac{-b-\sqrt{b^2-4ac}}{2a};\ x_2=\dfrac{-b+\sqrt{b^2-4ac}}{2a}[/tex]
[tex]r^2+4r-32\\\\a=1;\ b=4;\ c=-32\\\\b^2-4ac=4^2-4\cdot1\cdot(-32)=16+128=144\\\\\sqrt{b^2-4ac}=\sqrt{144}=12\\\\x_1=\dfrac{-4-12}{2\cdot1}=\dfrac{-16}{2}=-8\\\\x_2=\dfrac{-4+12}{2\cdot1}=\dfrac{8}{2}=4\\\\r^2+4r-32=(r-(-8))(r-4)=(r+8)(r-4)[/tex]
