SOLUTION: 25-x^2/12 times 6/x-5 I think I can cancel the 5 and 25 and the 12 and 6 5-x^2/2 times x-1/1 I can't seem to understand how to do this Can someone help.

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: 25-x^2/12 times 6/x-5 I think I can cancel the 5 and 25 and the 12 and 6 5-x^2/2 times x-1/1 I can't seem to understand how to do this Can someone help.      Log On


   

Question 130544: 25-x^2/12 times 6/x-5
I think I can cancel the 5 and 25 and the 12 and 6
5-x^2/2 times x-1/1
I can't seem to understand how to do this Can someone help.
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
%28%2825-x%5E2%29%2F12%29%286%2F%28x-5%29%29
I think I can cancel the 5 and 25 (No! No! A thousand times NO!) and the 12 and 6 (Yes)
%28%285-x%5E2%29%2F2%29%28%28x-1%29%2F1%29 (Your incorrect result)


Here's what you need to do:

%28%2825-x%5E2%29%2F12%29%286%2F%28x-5%29%29. Factoring a 6 out of the numerator of the second fraction and the denominator of the first was the right thing to do.

%28%2825-x%5E2%29%2F2%29%281%2F%28x-5%29%29. Now, recognize the numerator of the first fraction as the difference of 2 squares, a%5E2-b%5E2=%28a%2Bb%29%28a-b%29, so your expression becomes:

%28%28%285%2Bx%29%285-x%29%29%2F2%29%281%2F%28x-5%29%29. Now recognize that %285-x%29=-%28x-5%29. Your expression becomes:

%28%28-%285%2Bx%29%28x-5%29%29%2F2%29%281%2F%28x-5%29%29. NOW you can take the common factor of x-5 from the numerator and denominator

%28%28-%285%2Bx%29%2Across%28%28x-5%29%29%29%2F2%29%281%2Fcross%28%28x-5%29%29%29
%28%28-%285%2Bx%29%29%2F2%29%281%2F1%29
-%285%2Bx%29%2F2

Done.