Question

Perform the indicated operation and simplify your answer in factored form.

a^2+6a-7/2a^2+4a^2+2a*2a/a^2-1

Answer 1

Answer:

$\frac{(a-1)(a+7)(a-1)}{4a^{2}(a+1) }$

Step-by-step explanation:

The given expression is

$\frac{(a^{2} +6a-7)}{\frac{(2a^{3} +4a^{2} +2a)(2a)}{(a^{2} -1)} }$

First, we solve the division

$\frac{(a^{2} -1)(a^{2} +6a-7)}{(2a^{3} +4a^{2} +2a)(2a)}$

Second, we factor the numerator

$(a^{2} -1)(a^{2} +6a-7)=(a+1)(a-1)(a+7)(a-1)$

The first expression is the difference between two perfect squares, and the second expression is about finding to number which product is 7 and which difference is 6.

Third, we factor the denominator

$(2a^{3} +4a^{2} +2a)(2a)$

We extract the common factor from the trinomial

$2a(a^{2}+2a+1 )(2a)=4a^{2}(a^{2}+2a+1 )$

Now, we find two number which product is 1 and which sum is 2

$4a^{2}(a^{2}+2a+1 )=4a^{2}(a+1)(a+1)$

Then, we replace all factor in the initial fraction

$\frac{(a+1)(a-1)(a+7)(a-1)}{4a^{2}(a+1)(a+1) }$

Equal factors are simplified, given as result

$\frac{(a-1)(a+7)(a-1)}{4a^{2}(a+1) }$

Therefore, the answer in factored form is

$\frac{(a-1)(a+7)(a-1)}{4a^{2}(a+1) }$