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

Jul 14, 2016

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 $f \left(x\right) = q {x}^{n} + m {x}^{n - 1} + \ldots + p$. The roots are given by $\left(\text{factors of p")/("factors of q}\right)$.

In our case, the factors of $- 1$ are $\pm 1$ and the factors of $1$ are $\pm 1$. Hence, the rational root theorem says that our possible factors are $\frac{\pm 1}{\pm 1} = \pm 1$.

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 $f \left(x\right) = q {x}^{n} + m {x}^{n - 1} + \ldots + p$, then $f \left(a\right) = 0$.

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.

$\text{1_|1 1 -1 -1}$
$\text{ 1 2 1}$
"________"

$\text{ 1 2 1 0}$

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

${x}^{2} + 2 x + 1 = {\left(x + 1\right)}^{2}$.

So, ${x}^{2} + {x}^{2} - x - 1 = \left(x - 1\right) \left(x + 1\right) \left(x + 1\right) = \left(x - 1\right) {\left(x + 1\right)}^{2}$.

Hopefully this helps!