Question

Help please anyone!!!!!!

Answer 1

Answer:

$\huge\boxed{D)\frac{3p}{p-3} }$

Step-by-step explanation:

The first thing that can help you here is factoring every term.

$p^2-4p-12$

Factor.

$(p-6)(p+2)$

$6p+12$

Factor.

$6(p+2)$

18p

Factor, keeping in mind the denominator of the other term.

$6(3p)$

$p^2-9p+18$

Factor.

$(p-6)(p-3)$

Now you have simplified your original problem:

$\frac{p^2-4p-12}{6p+12} *\frac{18p}{p^2-9p+18}$

into:

$\frac{(p-6)(p+2)}{6(p+2)} *\frac{6(3p)}{(p-6)(p-3)}$

Now, after multiplying the numerator and denominator of the fractions:

$\frac{(p-6)(p+2)(6)(3p)}{(6)(p+2)(p-6)(p-3)}$

Then, you can simplify the fraction by cancelling terms:

$\frac{3p}{p-3}$

Hope it helps :) and let me know if you want me to elaborate.