Find the prime factorization of 945
1 answer:
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Answer:
The factors are: (3a+2b +ab-6)(3a+2b -ab+6)
Step-by-step explanation:
We need to solve the above expression using factorization.
Multiplying (a^2-4)(9-b^2)
9(a^2-4)-b^2(a^2-4) + 24ab
9a^2 -36 -a^2b^2+4b^2 + 24ab
Rearranging:
9a^2 + 4b^2 +24ab -36 -a^2b^2
We try to make perfect square of the form a^2+2ab-b^2
We have 24ab that can be written as 12ab + 12ab
Now, we can arrange the above equation:
9a^2 +12ab+ 4b^2 -(a^2b^2-12ab +36)
(3a)^2 +2(3a)(2b) + (2b)^2 -((ab)^2 -2(ab)(6)+(6)^2)
The perfect square will be:
(3a+2b)^2 - (ab-6)^2
Now We know a^2 - b^2 = (a+b)(a-b)
Here a = 3a+2b , b=ab-6
So,
(3a+2b +(ab-6))(3a+2b - (ab-6))
(3a+2b +ab-6)(3a+2b -ab+6)
So, the factors are: (3a+2b +ab-6)(3a+2b -ab+6)