Q:

# 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

Accepted Solution

A:
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 $$(x+2)(x-2)$$