Expressing quadratic equation in terms of its roots

Question

For a quadratic equation, $ax^2 + bx + c = 0$, why is
$ax^2 + bx + c = a(x-\alpha)(x-\beta)$ where alpha, beta are the roots of the equation? Why not just $(x-\alpha)(x-\beta)$?

Answer 1

In a quadratic equation where the coefficient of the first term ($x^2$) is unity, following holds:

1. the sum of the roots is equal to the coefficient of $x$ with its sign changed;
2. the product of the roots is equal to the third term.

Your equation is $\hspace{12 pt}ax^2+bx+c=0\hspace{12 pt}$coefficient of the first term ($x^2$) is NOT $1$.

If you write the equation as $\hspace{12 pt}a(x^2+\frac{b}{a}x+\frac{c}{a})=0$,

the coefficient of the first term of the quadratic expression in the bracket ($x^2$) is $1$.

Let the roots of the quadratic equation in the bracket be $\alpha,\beta$.

Then it can be written as $\hspace{12 pt}a(x-\alpha)(x-\beta)=0$,

where $\hspace{12 pt}\alpha+\beta=-\frac{b}{a}\hspace{12 pt}$ and $\hspace{12 pt}\alpha.\beta=\frac{c}{a}$