Please help me in factorising this equation

Question

$\ y^2-8xy+17x^2-4y+4=0 $

I have been trying to solve this factorisation but I am unable to reach the correct conclusion to this. Help on this would be highly appreciated.Thanks in advance

Answer 1

Note that $\ 2y^2-8xy+17x^2-4y+4=0$ is quadratic in $y$, in particular if we rewrite the equation as $$2y^{2} - (8x+4)y + 17x^{2}+4 = 0$$ we see that the equation is in the form $ay^{2}+by+c=0$ where $a = 2, b = -(8x+4),$ and $c = 17x^{2}+4$. Applying the quadratic formula gives us \begin{align*} y &= \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\\ &=\frac{8x+4\pm\sqrt{(8x+4)^{2}-4(2)(17x^{2}+4)}}{4}\\ &=\frac{8x+4\pm\sqrt{64x^{2}+64x+16-136x^{2}-32}}{4}\\ &=\frac{8x+4\pm\sqrt{-72x^{2}+64x-16}}{4}\\ &=\frac{8x+4\pm\sqrt{8(-9x^{2}+8x-2)}}{4}\\ &=\frac{8x+4\pm2\sqrt{2}\sqrt{-9x^{2}+8x-2}}{4}\\&=\frac{1}{2}\left(4x+2\pm\sqrt{2}\sqrt{-9x^{2}+8x-2}\right). \end{align*} This means that the equation can be factored as $$\left(y - \frac{1}{2}\left(4x+2+\sqrt{2}\sqrt{-9x^{2}+8x-2}\right)\right)\left(y-\frac{1}{2}\left(4x+2-\sqrt{2}\sqrt{-9x^{2}+8x-2}\right)\right)=0$$

If you instead think of the equation as a quadratic in $x$ you would get a similar result in the form $(x-r)(x-s) = 0$.

Answer 2

one possibility is $$\left(y-\frac{1}{2}\left(2+4x-\sqrt{2}\sqrt{-9x^2+8x-2}\right)\right)\left(y-\frac{1}{2}\left(2+4x+\sqrt{2}\sqrt{-9x^2+8x-2}\right)\right)$$

Answer 3

Assume that it could be factored. Then, by degree reasons, $$ 2y^2-8xy+17x^2-4y+4=(ax+by+c)(dx+ey+f) $$ This says that $ad=17$, $be=2$, $cf=4$ etc. We see that there is no solution in integers $a,b,c,d,e,f$. The same is true for $$ y^2-8xy+17x^2-4y+4. $$ But it can decomposed (not factored) as $(y-2)^2+(17x - 8y)x$.