[(a2 + 4a + 3)/(a2 - 8a + 7)] ÷ [(a2 + 8a +15)/(a2 - 9a + 8)] · [(a2 - 2a - 35)/(a2 - 7a - 8)]?

Answer 1

(a^2+4a+3)=(a+3)(a+1)
a^2-8a+7=(a-7)(a-1)
a^2+8a+15=(a+5)(a+3)
a^2-9a+8=(a-8)(a-1)
a^2-2a-35=(a-7)(a+5)
a^2-7a-8=(a-8)(a+1)
now plugging in
(a+3)(a+1)/(a-7)(a-1)divided by (a+5)(a+3)/(a-8)(a-1)*(a-7)(a+5)/(a-8)(a+1)
(a+3)(a+1)(a-8))(a-1)(a-7)(a+5)/(a+5)(a+3)(a-8)(a+1)
=(a-1)(a-7)

Answer 2

In order to do these types of problems you must be able to factor algebraic expressions. Do you know how? If not, you will never be successful simplifying these problems. You have to factor each trinomial, use the division rule for fractions and cancel the like factors in the numerator and denominator. In your problem, after factoring you should have

[(a + 1)(a + 3)]/[(a - 7)(a - 1)]*[(a - 8)(a - 1)]/[(a + 5)(a + 3)]*[(a - 7)(a + 5)]/[(a - 8)(a + 1)]

Cancel the a + 1, a - 7, a - 8, a + 3, a + 5 and a - 1. Final result is 1.

Answer 3

[(a+1)(a+3)/ (a-1)(a-7) ] :
[(a+3)(a+5)/(a-1)(a-8)].
[a-7)(a+5)/(a+1)(a-8)]

=1

Think which product of numbers gives the number without a.
factor and simplify

Answer: 1

Answer 4

the second response is right, but do you know how to get there....
factor flip the complex fraction over and multiply cross canceling then multiply again with the third equation, again cross canceling.