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

Question

Can somebody explain how to get from $(x^3 + x + 2)$ to $(x+1)\cdot(x^2-x+2)$ in several logical steps?

Answer 1

Note that $-1$ is a root of $x^3+x+2$, so by the factor theorem, $x+1$ is a factor of $x^3+x+2$.

Now, express $x^3+x+2 = (x+1)(ax^2+bx+c)$. Expand the right-hand side, and compare coefficients with the left-hand side.

Answer 2

A more or less "naïve" approach, without knowing the factor/residue theorem but knowing how to factor elementary quadratics:

$$x^3+x+2\stackrel{\color{red}{(*)}}=\color{green}{x^3}-x^2+\color{green}{x^2}+x+2=\color{green}{x^2(x+1)}-(x^2-x-2)=$$

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

The idea in $\;\color{red}{(*)}\;$ is to add and substract a term that will make factoring easier.

Answer 3

When we factor third degree polynomial, we must find at least one of its roots. This is because of degree: factors will be either of form $1+2$ or $1+1+1$ when it comes to degrees. Consequently, if there is no root in a field we want to factor the polynomial in, it is irreducible (this will never happen over $\Bbb C$ because of Fundamental theorem of algebra, but might over $\Bbb Q$).

Now, there is a very simple criterion for finding rational roots of polynomial with coefficients in $\Bbb Z$: rational root test. Thus, if $p/q$ is rational root of $x^3 + x +2$, then $p\mid2$ and $q\mid 1$, which means that if rational root exists, it is element of $\{\pm 1,\pm 2\}$. We can quickly eliminate positive candidates since $\alpha^3 + \alpha+2>0$ when $\alpha > 0$ and directly confirm that $-1$ is a root, while $-2$ is not.

Thus, we get that $x^3+x+2 = (x+1)(x^2+ax+b)$ and we can proceed in several ways. We could multiply $(x+1)(x^2+ax+b) = x^3 + (a+1)x^2+(a+b)x+b$ and compare coefficients to get $a = -1$, $b = 2$. We could use long division to directly compute $(x^3+x+2):(x+1)$. Or you could do something like DonAntonio did (being informed that $x+1$ is a factor): $$x^3+x+2 = x^3+x^2-x^2-x+2x+2 = x^2(x+1)-x(x+1)+2(x+1) = (x^2-x+2)(x+1)$$

Also, note that we already know that $x^2-x+2$ can't be factored further in $\Bbb Q$. If it could, we would already find its rational roots in the set $\{\pm 1,\pm 2\}$, but we've eliminated them early on.