Help understanding how to factor completely $x^3-x^2-x+1$

Question

I need someone to help explain the steps to completely factor the problem $x^3-x^2-x+1$.

Here is what I have done so far:

$x^3-x^2-x+1$

to

$x^3-x^2+-1(x+1)$

Since there is a negative I changed $-x+1$ to $-1(x+1)$

to

$x^2(x-1)+-1(x+1)$

to

$(x^2-1)(x-1)(x+1)$.

Is this correct? If not please explain what I did wrong and what I should do. If it is then thank you for saying so, if not, thank you for helping me to understand.

Answer 1

Let $f(x)=x^3-x^2-x+1$. By Rational Root Theorem, any rational root $\frac{p}{q}$ of $g(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ has $p\mid a_0$ and $q\mid a_n$ (i.e. $q$ divides $a_n$). Proof is simple: substitute $x=\frac{p}{q}$, multiply both sides by $q^n$, observe divisibility.

Since $f(1)=0$, we know $x^3-x^2-x+1=(x-1)P(x)$ for some $P(x)$, which of course must be quadratic, so $P(x)=x^2+ax+b$.

$$(x-1)(x^2+ax+b)=x^3+(a-1) x^2+(b-a)x-b=x^3-x^2-x+1$$

So $a-1=-1, b-a=-1, -b=1$, so $b=-1, a=0$.

$$x^3-x^2-x+1=(x-1)(x^2-1)$$

Answer 2

we want to factor the polynomial $\mathbb{P}=x^{3}-x^{2}-x+1$

factor out $x^2$ to start as follows; $\mathbb{P}=x^2(x-1)-(x-1)$

This makes sense because if you expand above you would get, $x^3-x^2-(x-1)=x^3-x^2-x+1$ which is of the form you had at the start.

Now factor out $(x-1)$

$$=(x-1)(x^2-1^2)$$

It may be helpful to see it as,

$(x-1)(x^2-1)$ say we call $(x-1)=a$ then you can see above on the second line for example we had $ax^2-a$. So to factor out a, can you see that it would be $a(x^2-1)$

I added the square because of the next step, and it does not change the one of course. Now, we know that $x^2-y^2=(x+y)(x-y)$ You can expand this to see that it is true if you like.

so $(x^2-1^2)=(x-1)(x+1)$

so $\mathbb{P}=(x-1)(x-1)(x+1)=(x-1)^2(x+1)$

Of course there are other ways that the polynomial can be factored, but this seems to be the nicest final form

Answer 3

$$x^3 - x^2 - x + 1 \to x^2(x - 1) + (-1)(x-1)$$ $$\to (x-1)(x^2 - 1) \to (x-1)^2(x+1)$$

Answer 4

\begin{align}x^3-x^2-x+1&=x^3-x^2-x+1+(3x^2-3x^2)+(3x-3x)\\ &=x^3+3x^2+3x+1-x^2-x-3x^2-3x\\ &=(x+1)^3-4x^2-4x\\ &=(x+1)^3-4x(x+1)\\ &=(x+1)\left[(x+1)^2-4x\right]\\ &=(x+1)(x-1)^2\end{align}

Answer 5

Simply notice that $x=1$ is a root. Hence by Ruffini your polynomial $x^3-x^2-x+1$ is divisible by $x-1$. Thus there exists a second degree polynomial $p(x)$ (you can find by a simple polynomial division) such that $x^3-x^2-x+1=(x-1)p(x)$. Being $p$ of degree $2$ you can easily find its roots with the well known $\Delta$ formula, in order to factorize completely your polynomial.