(r^2+7r+10)/3 * (3r-30)/(r^2-5r-50)

Question

How can i Factorize each expression

to end up with (r+5)(r+2)/3 * 3(r-10)/((r-10)(r+5)

can you explain.

Answer 1

(r+5)(r+2)(3)(r-10)/[ 3(r-10)(r+5)]

r+5 cancels
3 cancels
r-10 cancels
so
= (r+2)

Answer 2

I know that they cancel but what i want to know its how did this expression
(r^2+7r+10)/3 * (3r-30)/(r^2-5r-50)

became this (r+5)(r+2)/3 * 3(r-10)/((r-10)(r+5)

like the first expression had a 7r where is that in the second.

Answer 3

The 7r term disappears in the factoring process.

The second order polynomials in the numerator and denominator were both factored by Damon. You may need to review the factoring process; it is an important part of algebra. You can multiply the two monomial factors back together to verify that
(r-10)(r+5) = r^2 -5r -50,
for example.