Factoring the quartic polynomial $x^4-2x^2+1$

Question

Okay, I am practicing factoring for an upcoming assignment and I know that this is basic algebra, but I forgot how to attack this polynomial. Every method that I have used so far from simply guessing to using the quadratic formula to long division has failed me in replicating the answer. So either I am attacking this problem wrong up or there is another method that I just forgot.

So here it is: $x^4-2x^2+1=0\\$

The 4 roots of the polynomial are $\pm 1,\pm 1$ (it only has 2 distinct roots).

My confusion is in how to get those values.

Any help in deriving the solution would be greatly appreciated.

Answer 1

Hint:

Set $p=x^2$ to get $$p^2-2p+1=0$$

Solve this and then substitute back $p=x^2$.

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Even better would be if you note that $$x^4-2x^2+1=(x^2-1)^2$$

Answer 2

$x^4-2x^2+1=0$ clearly has a root of $1$. Since $\dfrac{x^4-2x^2+1}{x-1}= x^3+x^2-x-1$ you now have $$(x-1)(x^3+x^2-x-1)=0.$$

$x^3+x^2-x-1=0$ clearly has a root of $1$. Since $\dfrac{x^3+x^2-x-1}{x-1}= x^2+2x+1$ you now have $$(x-1)(x-1)(x^2+2x+1)=0.$$

And so on.

Answer 3

$$x^4-2x^2+1=0\\$$ $$\implies(x^2-1)^2 =0\\$$ $$\implies(x^2-1)(x^2-1)=0\\$$ $$\implies(x+1)(x-1)(x+1)(x-1)=0$$ $$\implies x=\pm1,\pm1$$