Question

How do you factor `(x^3+12x^2+48x+64)/8`?

Answer 1

`(x^3+12x^2+48x+64)/8 = 1/8(x+4)^3 = ((x+4)/2)^3`

Explanation:

Note that `x^3` and `64 = 4^3` are both perfect cubes.

Note also that:

`(a+b)^3 = a^3+3a^2b+3ab^2+b^3`

Putting `a=x` and `b=4`, we find:

`(x+4)^3 = x^3+3(4)x^2+3(4)^2x+4^3`

`color(white)((x+4)^3) = x^3+12x^2+48x+64`

So:

`(x^3+12x^2+48x+64)/8 = 1/8(x+4)^3 = ((x+4)/2)^3`

Answer 2

`(x(x^2+12x+48))/(8) +8`

Explanation:

To make the factoring easier, I'm going to rewrite the above expression as follows:

`(x^3+12x^2+48x)/8 + (64)/8`

NOTE: I separated the `64` so we could be able to factor the numerator.

We can factor an `x` out of the numerator, since all of the terms have an `x` in common. Factoring is essentially dividing every term by `x`, so we get:

`(x(x^2+12x+48))/(8) +8`

To factor the terms in parenthesis, let's do a small thought experiment:

• What two numbers when I add them, sum up to `12`, and when I multiply them, the product is `48?`

Well, the factors of `48` are `1` and `48`, `2` and `24`, `3` and `16`, `4` and `12`, and `6` and `8`. None of them add up to `12`, respectively. This means we cannot factor this any further with real numbers.