Perform indicated operation?

(2p)/(4p^2-1)/(6p^3)/(6p+3)2p4p2−1/6p36p+3

1 Answer
Roy Ø.
Jun 20, 2018

((2p)/(4p^2-1))/((6p^3)/(6p+3))2p4p2−16p36p+3 = 1/(p^2(2p-1))1p2(2p−1)

Explanation:

((2p)/(4p^2-1))/((6p^3)/(6p+3))2p4p2−16p36p+3

This would be the same as multiplying the numerator
(2p)/(4p^2-1)2p4p2−1
with the inverse of the denominator
(6p^3)/(6p+3)6p36p+3, i.e. (6p+3)/(6p^3)6p+36p3

We, therefore, get:
((2p)/(4p^2-1))/((6p^3)/(6p+3))2p4p2−16p36p+3
= ((2p)/(4p^2-1))*((6p+3)/(6p^3))(2p4p2−1)⋅(6p+36p3) = (2p(6p+3))/(6p^3(4p^2-1))2p(6p+3)6p3(4p2−1)
= (3(2p+1))/(3p^2(4p^2-1))3(2p+1)3p2(4p2−1) = ((2p+1))/(p^2(4p^2-1))(2p+1)p2(4p2−1)

Now we need to remember that (a+b)(a-b)=a^2-b^2
Therefore (2x+1)(2x-1)=(4x^2-1)

We can, therefore, write:
((2p+1))/(p^2(2p+1)(2p-1))=1/(p^2(2p-1))(2p+1)p2(2p+1)(2p−1)=1p2(2p−1)

Related questions
Impact of this question
4285 views around the world
You can reuse this answer
Creative Commons License