Question

Perform the indicated operation.

Answer 1

$\frac{2(a-3)}{(3+a)(3-a)}$

Step-by-step explanation:

This can be written as;

$\frac{8}{9-a^2} *\frac{a^2-6a+9}{4a^2-4a-24}$

Perform factorization

9-a²=difference of squares formula

(3+a)(3-a)

4a²-4a-24

4(a²-a-6)

4(a²-3a+2a-6)

4(a(a-3)+2(a-3))

4((a+2)(a-3))

a²-6a+9

a²-3a-3a+9

a(a-3)-3(a-3)

(a-3)(a-3)

Rewrite the operation as;

$\frac{8}{(3+a)(3-a)} *\frac{(a-3)(a-3)}{4((a+2)(a-3)}$

cancel common to remain with;

$\frac{2(a-3)}{(3+a)(3-a)}$

Learn More

Factorization : brainly.com/question/11930808

Keywords : Operation

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