Alissa Hancock

2022-07-07

A very simple question. How do we prove it?
$a{x}^{2}+bx+c=a\left(x-{r}_{1}\right)\left(x-{r}_{2}\right)$

Nathen Austin

Expert

$0=a{x}^{2}+bx+c=a{\left(x+\frac{b}{2a}\right)}^{2}-\frac{{b}^{2}-4ac}{4{a}^{2}},$
its roots are
${r}_{1,2}=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}.$
Depending on the sign of ${b}^{2}-4ac$, the roots could be distinct reals, a multiple real, or distinct complexes. We actually do not need the fundamental theorem of algebra here, which is usually used for equations of higher orders (e.g., order 5 or more).
Then we can evaluate it directly
$\begin{array}{rl}a\left(x-{r}_{1}\right)\left(x-{r}_{2}\right)& =a\left[{x}^{2}-\left({r}_{1}+{r}_{2}\right)x+{r}_{1}{r}_{2}\right]\\ & =a\left[{x}^{2}-\left(-\frac{b}{a}\right)x+\frac{{b}^{2}-\left({b}^{2}-4ac\right)}{4{a}^{2}}\right]\\ & =a\left({x}^{2}+\frac{b}{a}x+\frac{c}{a}\right)=a{x}^{2}+bx+c.\end{array}$