SOLUTION: Multipy. n^2-n-20/2n^2 x n^2 + 5n/n^2 - 25 note. the x between the fractions is a times sign. Thank you.

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Multipy. n^2-n-20/2n^2 x n^2 + 5n/n^2 - 25 note. the x between the fractions is a times sign. Thank you.      Log On


   

Question 29742: Multipy. n^2-n-20/2n^2 x n^2 + 5n/n^2 - 25
note. the x between the fractions is a times sign.
Thank you.
Answer by sdmmadam@yahoo.com(530) About Me  (Show Source):
You can put this solution on YOUR website!
n^2-n-20/2n^2 x n^2 + 5n/n^2 - 25
=[(n^2-n-20)/(2n^2)]X[(n^2 + 5n)/(n^2 - 25)]
=[(n-5)(n+4)/(2n^2)]X[n(n + 5)/(n+5)(n-5)]
Factorisation in the nr of the first fraction.
And using formula a^2-b^2 = (a+b)(a-b) and here a = n and b =5
in the dr of the second fraction
=[n(n-5)(n+4)(n+5)]/[(2n^2)(n+5)(n-5)]
=[(n+4)/(2n)] ( canceling (n-5), (n+5) and n)