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

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.

• Oct 25, 2014 at 19:54

Hint:

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

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

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

• I guess, I am just not seeing it. To me $p^2-2p+1=p(p-2)+1=0$ Oct 25, 2014 at 16:31
• That's true too, but this trinomial can also be factored into $(p-1)(p-1)$. This is called the product sum method. If you don't see this, feel free to use the quadratic formula, which gives the same answers, but is a bit more difficult in this case. Oct 25, 2014 at 16:35
• Right, simple method...$(p-1)(p-1)=0 \implies (p-1)^2=0 \implies (x^2-1)^2=0\\$ Then $(x^2-1)=0 \implies x^2=1 \implies \sqrt{x^2}=\sqrt{1} \implies x=\pm \sqrt{1} = \pm 1$ Then repeat for the other term. Giving 4 roots of $\pm 1$ Oct 25, 2014 at 16:48
• Your method works fine, but I don't understand what you say after the math. $x^4-2x^2+1$ only has 2 roots, which are $x=\pm1$. Oct 25, 2014 at 16:49
• It must have 4 roots though if it is of degree 4, no? The solution to the problem states: $x=\pm 1, \pm 1$ Implying 4 real roots. So if we have from above $(x^2-1)^2=(x^2-1)(x^2-1)$ and thus we have two terms in which to solve the roots for. No? $(x^2-1)=0 \implies x=\pm 1$ and then again $(x^2-1)=0 \implies x=\pm 1$ Oct 25, 2014 at 16:55

$x^4-2x^2+1=0$ 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.

• So use long division... Oct 25, 2014 at 16:35
• As someone who struggles to remember the rules for factoring trinomials and greater, it's important to be good at guessing one root and doing long division.
$$x^4-2x^2+1=0\\$$ $$\implies(x^2-1)^2 =0\\$$ $$\implies(x^2-1)(x^2-1)=0\\$$ $$\implies(x+1)(x-1)(x+1)(x-1)=0$$ $$\implies x=\pm1,\pm1$$