[tex]\frac{2(a-3)}{(3+a)(3-a)}[/tex]
Step-by-step explanation:
This can be written as;
[tex]\frac{8}{9-a^2} *\frac{a^2-6a+9}{4a^2-4a-24}[/tex]
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;
[tex]\frac{8}{(3+a)(3-a)} *\frac{(a-3)(a-3)}{4((a+2)(a-3)}[/tex]
cancel common to remain with;
[tex]\frac{2(a-3)}{(3+a)(3-a)}[/tex]
Learn More
Factorization : https://brainly.com/question/11930808
Keywords : Operation
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