Question

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

Answer 1

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 two numbers that will multiply to be equal the c, and will also add up to equal b. Those numbers are 2 and 2:
$x^{2} +4x+4=(x+2)(x+2)$
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 $(x+2)(x-2)$