Factorize this polynomial $ax^2+bx+c$ into factors of the first exponent in the cases when D>0, D=0

Question

The previous request was to prove the identity $ax^2+bx+c=a[(x+(b/2a)^2-(D/4a^2)]$, where $D=b^2-4ac$
And I proved it from the left to the right, which means I managed to express $ax^2+bx+c$ as $a[(x+(b/2a)^2-(D/4a^2)]$.
What should I do next?

Answer 1

Since your original goal is to factorise, using $a^2-b^2 = (a+b)(a-b)$, $$\begin{align*} ax^2+bx+c &= a\left[\left(x+\frac b{2a}\right)^2 - \frac D{4a^2}\right]\\ &= a\left[\left(x+\frac b{2a}\right)^2 - \left(\frac{\sqrt D}{2a}\right)^2\right]\\ &= a\left(x+\frac b{2a}+\frac{\sqrt D}{2a}\right)\left(x+\frac b{2a}-\frac{\sqrt D}{2a}\right)\\ &= a\left(x-\frac {-b-\sqrt D}{2a}\right)\left(x-\frac {-b+\sqrt D}{2a}\right)\\ \end{align*}$$ This is a more general case where $D\ge 0$. If $D=0$, then the two factors are the same, i.e. have a second power: $$ax^2+bx+c = a\left(x-\frac {-b}{2a}\right)^2$$

Answer 2

If $D = 0$, then you have $$ax^2 + bx + c = a\left(x + \frac b{2a}\right)^2$$ where $c = \left(\dfrac b{2a}\right)^2 = \dfrac{b^2}{4a^2}$.

If $D> 0$, then you have the identity you've proven. $$ax^2+bx+c=a[(x+(b/2a)^2-(D/4a^2)] = a\left(x-\frac {-b-\sqrt D}{2a}\right)\left(x-\frac {-b+\sqrt D}{2a}\right)$$

The right side follows because you have a difference of squares of the form $x^2 -y^2 = (x-y)(x+y)$.