Divide the following fractions.6p-18 over 9p divide by 3p-9 over p^2+2p?

Answer 1

This time, start by changing the problem to multiplication. Leave the first fraction the same, but reciprocate the second fraction.

(6p - 18)/(9p) * (p² + 2p)/(3p - 9)

Again, factor all the polynomials.

6p - 18 = 6(p - 3)

p² + 2p = p(p + 2)

9p cannot be factored

3p - 9 = 3(p - 3)

So, the numerator is 6(p - 3)p(p + 2)
and the denominator is 9p*3(p - 3)
the (p - 3) cancels (p - 3), the p cancels p.
This leaves you with 6(p + 2)/27.

Finally, the 6 and 27 reduce to 2/9

This leaves 2(p + 2)/9 or (2p + 4)/9

Answer 2

{6(p-3)/9p } * {p(p+2)/3(p-3)}
=2(p+2)/9