Find $y$ in $y^4-6y^3+14y^2-20y+8=0$

Question

Find $y$ in $$\displaystyle y \cdot \frac{6-y}{y+1}\bigg(\frac{6-y}{y+1}+y\bigg) = 8$$

Solution I tried $y(6-y)(6+y^2)=8(y+1)^2$

$(6y-y^2)(6+y^2)=8(y^2+2y+1)$

$(36y+6y^3-6y^2-y^4)=8y^2+16y+8$

$y^4-6y^3+14y^2-20y+8=0$

Help me how to solve it after that point.

Answer 1

You could start this way. Let $z=\cfrac {6-y}{y+1}$

Then, clearing fractions $yz=6-y-z$.

You also have $yz(y+z)=8$ so letting $t=y+z$ you get the quadratic $(6-t)t=8$ which you can solve.

Then you have two values of $t$ and each yields a quadratic for $y$.

Answer 2

It factorises as $$(y^2-4y+2)(y^2-2y+4)=0,$$ so the solutions are $y=2\pm\sqrt{2}, 1\pm i\sqrt{3}$.

Answer 3

make the Ansatz $$(y^2+ay+b)(y^2+cy+d)=y^4-6y^3+14y^2-20y+8$$ for your control it is $$(2-4y+y^2)(4-2y+y^2)$$

Answer 4

For all real $k$ we have: $$y^4-6y^3+14y^2-20y+8=(y^2-3y+k)^2-9y^2-2ky^2+6ky-k^2+14y^2-20y+8=$$ $$=(y^2-3y+k)^2-((2k-5)y^2-(6k-20)y+k^2-8)=$$ $$=(y^2-3k+3)^2-(y^2+2y+1)^2=(y^2-3y+3)^2-(y+1)^2=(y^2-4y+2)(y^2-2y+4).$$ Can you end it now?