Question

How do you factor `x^3+x^2-x-1`?

Answer 1

We must first determine what numbers are factors using the rational root theorem and the remainder theorem.

Explanation:

The rational root theorem helps us list all possible factors of a particular polynomial of the form `f(x) = qx^n + mx^(n - 1) + ... + p`. The roots are given by `("factors of p")/("factors of q")`.

In our case, the factors of `-1` are `+-1` and the factors of `1` are `+-1`. Hence, the rational root theorem says that our possible factors are `(+-1)/(+-1) = +-1`.

The remainder theorem helps to see if a given number is a factor without having to effectuate the full division. If `x - a` is a factor of `f(x) = qx^n + mx^(n - 1) + ... + p`, then `f(a) = 0`.

Let's start with the positive `1`.

`1^3 + 1^2 - 1 - 1 =^? 0`

`2 - 2 = 0`

Yes, `x - 1` is a factor of `x^3 + x^2 - x - 1`!

Now, we can use either synthetic or long division to start the factoring process. I will use synthetic, because for me it is extremely efficient and quick at getting the job done.

`"1_|1 1 -1 -1"`
`" 1 2 1"`
"________"

`" 1 2 1 0"`

Hence, `(x^3 + x^2 - x - 1) ÷ (x - 1) = x^2 + 2x + 1`. Otherwise put `(x - 1)(x^2 + 2x + 1) = x^3 + x^2 - x - 1`. We're not done yet, though. `x^2 + 2x + 1` can be factored as a perfect square trinomial, `(a + b)(a + b) = (a + b)^2`.

`x^2 + 2x + 1 = (x + 1)^2`.

So, `x^2 + x^2 - x - 1 = (x - 1)(x + 1)(x + 1) = (x - 1)(x + 1)^2`.

Hopefully this helps!