Why are $(x-x_1)$ and $(x-x_2)$, where $x_1,x_2$ are roots of a quadratic equation, factors of said equation?

Question

I can not figure it out. Thanks.

Answer 1

Suppose we have the equation:

$ax^{2}+bx+c=0$.

The quadratic formula states:

$x=\frac{-b+\sqrt{b^{2}-4ac}}{2a}$ and $x=\frac{-b-\sqrt{b^{2}-4ac}}{2a}$ are the roots of this quadratic equation.

Since:

$0=(x-(\frac{-b+\sqrt{b^{2}-4ac}}{2a}))(x-(\frac{-b-\sqrt{b^{2}-4ac}}{2a}))=x^{2}-x(\frac{-b+\sqrt{b^{2}-4ac}+(-b)-\sqrt{b^{2}-4ac}}{2a})+\frac{c}{a}=ax^{2}+bx+c$

It follows that for any quadratic polynomial, its roots are factors.

Answer 2

A more general argument that applies to roots of polynomials of arbitrary degree is as follows.

The substitution mapping $\phi_a: \mathbb{F}[X] \to \mathbb{F}$ defined by $f \mapsto f(a)$ is a homomorphism, thus its kernel is an ideal of $\mathbb{F}[X]$. We know that $(X - a) \subset ker(\phi_a)$ and so the ideal $(X - a)$ must in fact be the whole kernel since it is maximal. Therefore any polynomial that has $a$ as a root must be divisible by $X-a$.