Question

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

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)`