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


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

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

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

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