Answer:
[tex]\huge\boxed{D)\frac{3p}{p-3} }[/tex]
Step-by-step explanation:
The first thing that can help you here is factoring every term.
[tex]p^2-4p-12[/tex]
Factor.
[tex](p-6)(p+2)[/tex]
[tex]6p+12[/tex]
Factor.
[tex]6(p+2)[/tex]
18p
Factor, keeping in mind the denominator of the other term.
[tex]6(3p)[/tex]
[tex]p^2-9p+18[/tex]
Factor.
[tex](p-6)(p-3)[/tex]
Now you have simplified your original problem:
[tex]\frac{p^2-4p-12}{6p+12} *\frac{18p}{p^2-9p+18}[/tex]
into:
[tex]\frac{(p-6)(p+2)}{6(p+2)} *\frac{6(3p)}{(p-6)(p-3)}[/tex]
Now, after multiplying the numerator and denominator of the fractions:
[tex]\frac{(p-6)(p+2)(6)(3p)}{(6)(p+2)(p-6)(p-3)}[/tex]
Then, you can simplify the fraction by cancelling terms:
[tex]\frac{3p}{p-3}[/tex]
Hope it helps :) and let me know if you want me to elaborate.
