Find roots of $x^4 -6x^3 + 12x^2 - 12 x + 4 = 0$

Question

the original equation is:

$$(x^2 + 2)^2 -6x(x^2+2) + 8x^2=0.$$

cannot see how to go solving this. I tried following way to factorise:

$$(x^2+2)(x^2-6x+2) + 8x^2 = 0.$$

But this has no help to solve.

Thank you people, but I need the thinking process, not the answer.

Answer 1

It's $$\left(\frac{x^2+2}{x}\right)^2-6\left(\frac{x^2+2}{x}\right)+8=0,$$ which gives $$\frac{x^2+2}{x}=2$$ and $x\in\{1+i,1-i\}$ or $$\frac{x^2+2}{x}=4,$$ which gives $x\in\{2+\sqrt2,2-\sqrt2\}$.

Answer 2

In the original version, let $y=x^2+2$. Then, we have $y^2-6xy+8x^2=0$, which gives us $(y-2x)(y-4x)=0$.

Answer 3

your equation is equivalent to $$(x^2-4x+2)(x^2-2x+2)=0$$

Answer 4

One way to find the factorizations others have given is to complete the square. If you recognize $y^2-6y+9=(y-3)^2$ you can do $(x^2+2)^2-6(x^2+2)+8x^2=((x^2+2)-3x)^2-x^2=(x^2+2-4x)(x^2+2-2x)$

Answer 5

Suppose, instead that the problem was $$(x^2+2)^2 -6t(x^2+2)+8t^2=0$$

Now suppose that $x=t$