Hi;
If a quadratic polynomial won't factor does this mean it
has imaginary roots?
if I complete the square on the same polynomial which makes it
factorable does this process give it real roots?
If it won't factor over the reals, then yes, it will have COMPLEX roots (which have a real part and an imaginary part).
Completing the square won't magically make a quadratic factorise over the reals. You WILL however be able to factorise over the complex numbers, which will still give complex roots.if I complete the square on the same polynomial which makes it
factorable does this process give it real roots?
The point is that any quadratic with real coefficients can be factored.
Take the example I gave you. $\displaystyle 3x^2-3x-2=0$ has two real roots: $\displaystyle \left( {\frac{{3 + \sqrt {33} }}{6}} \right)\;\& \;\left( {\frac{{3 - \sqrt {33} }}{6}} \right)$.
The factored form is: $\displaystyle 3x^2-3x-2=\left[ {x - \left( {\frac{{3 + \sqrt {33} }}{6}} \right)} \right]\left[ {x - \left( {\frac{{3 - \sqrt {33} }}{6}} \right)} \right]=0$.
Any quadratic, whether its coefficients are real or not, can be factored- but NOT necessarily with integer or even real coefficients.
Take $\displaystyle ax^2+ bx+ c$ as a generic example. First, factor out the "a". $\displaystyle a(x^2+ \frac{b}{a}x+ \frac{c}{a})$. Now "complete the square"- add and subtract $\displaystyle \frac{b^2}{4a^2}$: $\displaystyle a(x^2+ \frac{b}{a}x+ \frac{b^2}{4a}+ \frac{c}{a}- \frac{b^2}{4a^2})$. The point of that is that $\displaystyle x^2+ \frac{b}{a}x+ \frac{b^2}{4a^2}= (x+ \frac{b}{2a})^2$, a perfect square. And of course, $\displaystyle \frac{b^2}{4ac^2}-\frac{c}{a}= \left(\sqrt{\frac{b^2}{4a^2}- \frac{c}{a}}\right)^2$.
That is, we can write $\displaystyle ax^2+ bx+ c= a\left(\left(x+ \frac{b}{2a}\right)^2- \left(\sqrt{\frac{b^2}{4a^2}- \frac{c}{a}}\right)^2\right)$
And now we use the fact that $\displaystyle x^2- a^2= (x- a)(x+ a)$ to write
$\displaystyle a\left(x+ \frac{b}{2a}+\sqrt{\frac{b^2}{4a^2}- \frac{c}{a}}\right)\left(x+ \frac{b}{2a}- \sqrt{\frac{b^2}{4a^2}- \frac{c}{a}}\right)$
That is factored into a number times two linear factors. With those square roots, those numbers may not be integers or even real numbers.