# factoring a quadratic over $\mathbb{C}$

someone please help me, I have tried so many times to factor this basic polynomial over $$\mathbb{C}$$. its

$$ax^2+bxy+cy^2$$

I saw someone answered a question and Im sorry I had to ask this again because they just brushed over it and I cannot see why that user factored out by $$\frac{1}{4a}$$ This is frustrating because my original problem was to show a singular conic is the union of two lines and I did all the actual work but cannot do this advanced high school level factoring part

Again sorry if this is too easy or if someone did it, but I cannot sleep I fully understand every move I make in every proof and solution

Edit: I tried using quadratic formula is that the way to go? Just wasn't getting back the $$a$$ in my first term!

I get $$(x-y(\frac{-b+\sqrt{b^2-4ac}}{2a}))(x-y(\frac{-b-\sqrt{b^2-4ac}}{2a}))$$

Look at $$ax^2 + bxy +cy^2$$ as a quadratic in $$x$$ seeing $$y$$ as a constant to express $$x$$ in terms of $$y$$. The determinant is $$(by)^2 - 4a(cy^2)= b^2 y^2 - 4ac y^2 = y^2(b^2-4ac)$$ and then we have cases depending on whether $$b^2-4ac <0$$ or not. If positive, we get the two solutions

$$x= \frac{-by + y\sqrt{b^2-4ac}}{2a} , x= \frac{-by - y\sqrt{b^2-4ac}}{2a}$$

so that we get the factorisation

$$a\left(x-y\frac{-b + \sqrt{b^2-4ac}}{2a}\right)\left(x+y\frac{-b + \sqrt{b^2-4ac}}{2a}\right)$$

and for negative determinant

$$a\left(x-iy\frac{-b + \sqrt{b^2-4ac}}{2a}\right)\left(x+iy\frac{-b + \sqrt{b^2-4ac}}{2a}\right)$$

• ya I realized I needed to just multiply by $a$ and I was done lol thanks@ – Hossien Sahebjame Mar 7 '20 at 23:22