SOLUTION: Hello tutors! I'm stuck on this one. The answer is 4a^2/[9(a-1)], but I'm not sure how the 5 factors out. Thanks for any help you can give! 5a^2-20 ( a^3+2a^2 9a^3+6a^2) _

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Hello tutors! I'm stuck on this one. The answer is 4a^2/[9(a-1)], but I'm not sure how the 5 factors out. Thanks for any help you can give! 5a^2-20 ( a^3+2a^2 9a^3+6a^2) _      Log On


   

Question 87264: Hello tutors! I'm stuck on this one. The answer is 4a^2/[9(a-1)], but I'm not sure how the 5 factors out. Thanks for any help you can give!
5a^2-20 ( a^3+2a^2 9a^3+6a^2)
_______ / ________ * __________
3a^2-12a (2a^2-8a 2a^2-4a)
Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
You are looking for factorised versions that have same factors in so that we can cancel down and thus simplify.

i will do the numerator first:
+%285a%5E2-20%29%2F%283a%5E2-12a%29++
+%285%28a%5E2-4%29%29%2F%283a%28a-4%29%29++
+%285%28a%2B2%29%28a-2%29%29%2F%283a%28a-4%29%29++

And now the denominator --> has multiplications and divisions so it shouldn't matter which order we do them: I shall factorise everything first to see what i have.




+%28+3a%5E4%283a%2B2%29%28a%2B2%29+%29+%2F+%28+4a%5E2%28a-4%29%28a-2%29+%29+
+%28+3a%5E2%283a%2B2%29%28a%2B2%29+%29+%2F+%28+4%28a-4%29%28a-2%29+%29+

what we have left now looks just as complicated but that is only because we have so many fractions in there.

Consider +%28a%2Fb%29%2F%28c%2Fd%29+. This can be re-written as +%28a%2Fb%29+%2A+%28d%2Fc%29+

so our can be thought of as





and finally we can start to simplify by cancelling out terms.
+%28+%2820%28a-2%29%28a-4%29%28a-2%29%29%2F%289a%5E3%28a-4%29%283a%2B2%29%29+%29+
+%28+%2820%28a-2%29%28a-2%29%29%2F%289a%5E3%283a%2B2%29%29+%29+

which is nowhere near what you say is the answer so i am guessing that my view of what you wrote is wrong... my method is fine - i have checked it a couple of times but admittedly it is difficult on here rather than on paper.

See if you follow what i did and apply it to your version of the question and see if you get the answer you quote.

Good luck
cheers
Jon.