How do you multiply (f^2 - 7f + 10)/( f^3 - 7f^2 + 7f - 49) *( f^2 - 49)/( 2f^3-10*f^2) ?

1 Answer
Sep 18, 2015

(f^2 - 7f + 10)/( f^3 - 7f^2 + 7f - 49) *( f^2 - 49)/( 2f^3-10*f^2) = ((f-2)(f+7))/((f^2+7)(2f^2))

Explanation:

We know that f^2-49 is a difference of squares which can also be written as (f-7)(f+7)

2f^3 - 10f^2 has two factors we can put in evidence, 2 and f^2, so we can rewrite it as 2f^2(f-5)

f^3 - 7f^2 + 7f - 49 is a bit more tricky, but you can get out by grouping by parts. To do this let's put f^2 in evidence whenever we can, so we have f^2(f - 7) + 7f - 49

Now the two terms left have 7 as a common term, so we can put that in evidence and give us, f^2(f-7) +7(f-7). Now we have (f-7) in both terms so we can put that in evidence, giving us the factored form: (f^2+7)(f-7)

f^2 - 7f + 10 is a quadratic so we can just use normal factoring methods for that:
Since the f^2 coefficient is 1 we know the factors will be of the form (x+a)(x+b). Now, we know that ab = 10, and the two obvious names to come to mind are 2 and 5. We know that they must sum out to be the f coefficient times -1 so 2 and 5 fit. So now we put it all together and cancel everything that shows up on both numerator and denominator

((f-2)cancel((f-5)))/((f^2+7)cancel((f-7))) *(cancel((f-7))(f+7))/(2f^2(cancel(f-5))) = ((f-2)(f+7))/((f^2+7)(2f^2))

Factoring functions like this can be a bit hard, specially for non-quadratics, but I've found that this is a good source..