# Solving a quartic polynomial

#### anemone

##### MHB POTW Director
Staff member
Solve $$\displaystyle x^4+1=2x(x^2+1)$$.

#### MarkFL

Staff member
I would first write the quartic in standard form and assume it has two quadratic factors:

$$\displaystyle x^4-2x^3-2x+1=(x^2+ax+1)(x^2+bx+1)=x^4+(a+b)x^3+(2+ab)x^2+(a+b)x+1$$

Equating coefficients, we find:

$$\displaystyle a+b=-2$$

$$\displaystyle 2+ab=0$$

We have two solutions, but they are symmetric, hence we may take:

$$\displaystyle a=\sqrt{3}-1,\,b=-(\sqrt{3}+1)$$

Hence:

$$\displaystyle x^4-2x^3-2x+1=(x^2+(\sqrt{3}-1)x+1)(x^2-(\sqrt{3}+1)x+1)=0$$

Application of the quadratic formula on the two factors gives us:

$$\displaystyle x=\frac{1-\sqrt{3}\pm i\sqrt{2\sqrt{3}}}{2},\,\frac{1+\sqrt{3}\pm\sqrt{2\sqrt{3}}}{2}$$

#### topsquark

##### Well-known member
MHB Math Helper
You could try this. But I'll admit that I've never gotten through one of those without making any mistakes. It is, in a word, "tedious."

-Dan

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#### anemone

##### MHB POTW Director
Staff member
You could try this. But I'll admit that I've never gotten through one of those without making any mistakes. It is, in a word, "tedious."

-Dan
Hi Dan,

The link doesn't work... And my solution is less clever than the approach of MarkFL. First, I let $$\displaystyle x=\tan \theta$$ to transform the original equation and making it as $$\displaystyle \sin^2 2\theta+2\sin 2\theta-2=0$$.

I then use the quadratic formula to solve for $$\displaystyle \sin 2\theta$$ then $$\displaystyle \tan \theta$$ to obtain the numerical approximation of the answers where

$$\displaystyle x=\tan 23.5293^{\circ}=0.4354$$ and $$\displaystyle x=\tan 66.4707^{\circ}=2.2966$$.

Also, I was not able to find the complex roots as well!   #### topsquark

##### Well-known member
MHB Math Helper
Hi Dan,

The link doesn't work... It should work now. Thanks for the catch.

-Dan