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pogonyaev
3 years ago
9

Part A: Factor x2b2 − xb2 − 6b2. Show your work. (4 points)

Mathematics
1 answer:
IrinaVladis [17]3 years ago
7 0
PART A. Notice that we have b^{2} as a common factor in all the terms, so lets factor that out:
x^{2} b^2-xb^2-6b^2
b^2(x^{2} -x-6)
Now we need can factor x^{2} -x-6:
b^2(x^{2} -x-6)
b^2(x+2)(x-3)

We can conclude that the complete factorization of x^{2} b^2-xb^2-6b^2 is b^2(x+2)(x-3).

PART 2. Here we just have a quadratic expression of the form a x^{2} +bx+c. To factor it, we are going to find <span>two numbers that will multiply to be equal the </span>c<span>, and will also add up to equal </span><span>b. Those numbers are 2 and 2:
</span>x^{2} +4x+4=(x+2)(x+2)
Since both factors are equal, we can factor the expression even more:
x^{2} +4x+4=(x+2)^2

We can conclude that the complete factorization of x^{2} +4x+4 is (x+2)^2.

PART C. Here we have a difference of squares. Notice that 4, can be written as 2^2, so we can rewrite our expression:
x^2-4=x^2-2^2
Now we can factor our difference of squares like follows:
x^{2} -2^2=(x+2)(x-2)

We can conclude that the complete factorization of x^2-4 is (x+2)(x-2)
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