1. ## How do you factor an un-prime quadratic?

like 6x^2+2x-5=0

I can find solutions but they are generally wrong like (2x-5)(6x+2)=0.

I think polynomial long-division might work, come to think of it.

2. Here's a kick-off

(ax+b)(cx+d) where ac=6 and bd=-5

now expand (ax+b)(cx+d) out, what do you get?

3. Originally Posted by bournouli
like 6x^2+2x-5=0

I can find solutions but they are generally wrong like (2x-5)(6x+2)=0.

I think polynomial long-division might work, come to think of it.
Since the discriminant is irrational it will not factor with rational coefficients. Here is a way it can be done.

$\displaystyle 6x^2+2x-5=0 \iff 6\left[x^2+\frac{1}{3}x-\frac{5}{6}\right]=0$

Now complete the square to get

$\displaystyle 6\left[ \left(x^2+\frac{1}{3}x+\frac{1}{36}\right)-\frac{1}{36}-\frac{5}{6}\right]=6\left[ \left(x+\frac{1}{6} \right)^2-\frac{31}{36}\right]$

Now we can write this as the difference of squares

$\displaystyle 6\left[ \left(x+\frac{1}{6} \right)^2-\left(\frac{\sqrt{31}}{6}\right)^2\right]=6\left[ \left(x+\frac{1}{6} -\frac{\sqrt{31}}{6}\right)\left(x+\frac{1}{6} +\frac{\sqrt{31}}{6}\right)\right]$

4. It was really that easy huh? -thanks.