Part A: Factor x2b2 − xb2 − 6b2. Show your work. (4 points)Part B: Factor x2 + 4x + 4. Show your work. (3 points)Part C: Factor x2 − 4. Show your work. (3 points)Please help im not the best with math

PART A. Notice that we have [tex]b^{2}[/tex] as a common factor in all the terms, so lets factor that out:
[tex] x^{2} b^2-xb^2-6b^2[/tex]
[tex]b^2(x^{2} -x-6)[/tex]
Now we need can factor [tex]x^{2} -x-6[/tex]:
[tex]b^2(x^{2} -x-6)[/tex]
[tex]b^2(x+2)(x-3)[/tex]

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

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

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

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

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