# Possible "clever" ways to solve $x^4+x^3-2x+1=0$, with methodological justification

Solve the quartic polynomial : $$x^4+x^3-2x+1=0$$ where $$x\in\Bbb C$$.

Algebraic, trigonometric and all possible methods are allowed.

I am aware that, there exist a general quartic formula. (Ferrari's formula). But, the author says, this equation doesn't require general formula. We need some substitutions here.

I realized there is no any rational root, by the rational root theorem.

The harder part is, WolframAlpha says the factorisation over $$\Bbb Q$$ is impossible.

Another solution method can be considered as the quasi-symmetric equations approach. (divide by $$x^2$$).

$$x^2+\frac 1{x^2}+x-\frac 2x=0$$

But the substitution $$z=x+\frac 1x$$ doesn't make any sense.

I want to ask the question here to find possible smarter ways to solve the quartic.

• The only irreducibles over $\mathbb{R}$ are linear or quadratic. Perhaps this is irreducible over $\mathbb{Q}$? Oct 4, 2022 at 23:33
• Factorization over $\mathbb R$ is always possible. Here you get two quadratic factors. Oct 4, 2022 at 23:34
• Have you tried replacing $x$ by something like $x-1$ and using Eisenstein’s criterion? Oct 4, 2022 at 23:35
• Asking WA to find the roots reveals that this polynomial factors into a pair of irreducible quadratics over $\mathbb{Q}(\sqrt{-3})$. I don't see a nice way of seeing this from the equation alone, though. Computing the resolvent cubic could be helpful. Oct 4, 2022 at 23:43
• @MichaelBurr Yes, I meant over $\Bbb Q$
– User
Oct 5, 2022 at 0:19

Substitute $$x=y+1$$, then $$z=y+1/y$$ to get $$y^4+5y^3+9y^2+5y+1=0$$ $$y^2((y+1/y)^2+5(y+1/y)+7)=0$$ $$z^2+5z+7=0$$ Then $$z=\frac{-5\pm\sqrt{-3}}2$$ and $$x=\frac{z\pm\sqrt{z^2-4}}2+1$$, where the four solutions are obtained from the different sign choices for $$z$$ and $$x$$.

More concretely, let $$t,u=\frac{-1\pm\sqrt{-3}}2$$ be the two roots of $$x^2+x+1$$, then $$x^4+x^3-2x+1=(x^2-tx+t)(x^2-ux+u)$$

• Hello. How did you find this substitution?
– User
Oct 5, 2022 at 0:25
• @User Just try linear shifts. Oct 5, 2022 at 0:40
• You factorize into quadratric polynomials with complex coefficients; they have an "elegant" form and are of course complex conjugate. One can use your solutions for $x$ to factorize into quadratric polynomials with real coefficients; but these are fairly "ugly". Oct 7, 2022 at 16:51

We can look for a difference of squares factorization. Completing the square gives

$$\left( x^2 + \frac{1}{2} x + c \right)^2 - \left( 2c + \frac{1}{4} \right) x^2 - (c + 2) x - (c^2 - 1)$$

and we want to find a value of $$c$$ such that the discriminant of the quadratic on the right is equal to zero. This gives

$$\Delta = (c + 2)^2 - 4 \left( 2c + \frac{1}{4} \right) \left( c^2 - 1) \right) = - 8c^3 + 12c + 5$$

which happily has a rational root $$c = - \frac{1}{2}$$ (I guess we must be essentially using the resolvent cubic here). This gives us a factorization

$$\left( x^2 + \frac{1}{2} x - \frac{1}{2} \right)^2 + \frac{3}{4} (x - 1)^2$$

which gives a difference of squares factorization

$$\left( x^2 + \frac{1}{2} x - \frac{1}{2} + \frac{i \sqrt{3}}{2} (x - 1) \right) \left( x^2 + \frac{1}{2} x - \frac{1}{2} - \frac{i \sqrt{3}}{2} (x - 1) \right)$$

and we can use the quadratic formula from here; if you want to know what the roots end up looking like you can ask WolframAlpha.

• Hello. Is the factorisation over $\Bbb R$ is always possible for quartic?
– User
Oct 5, 2022 at 0:24
• Yes, it’s always possible. The complex roots are either real or come in conjugate pairs. Oct 5, 2022 at 0:25
• Thanks. Factorisation over $\Bbb R$, means $x^4+bx^3+...+e=(x^2+mx+n)(x^2+px+q)$ is always possible, where $m,n,p,q\in\Bbb R$. Is my reasoning correct?
– User
Oct 5, 2022 at 0:30
• Yes, that’s correct. More generally any polynomial over the reals factors into a product of linear and quadratic polynomials, and this is equivalent to the fundamental theorem of algebra. Oct 5, 2022 at 0:31
• @User That's also intuitively the reason why adding to $\mathbb{R}$ a solution to $X^2+1=0$ (that's $i$) is sufficient to split all polynomials into degree 1 polynomials: the only polynomials of degree $>1$ in $\mathbb{R}[X]$ that cannot be split into a product of smaller degree polynomials have degree 2. Oct 9, 2022 at 16:41

Since this quartic has no real roots, it has two pairs of complex conjugate roots, so it must factor into two conjugate quadratics:

$$(x^2 + ax + b)(x^2 + \overline ax + \overline b) = x^4 + (a + \overline a)x^3 + (a\overline a + b + \overline b)x^2 + (a\overline b + \overline ab)x + b\overline b.$$

The $$x^3$$ coefficient is $$1 = a + \overline a$$, so we must have

$$a = \frac12 + si$$

for some $$s ∈ ℝ$$. The constant term is $$1 = b\overline b$$, so $$b$$ lies on the complex unit circle, and we must have

$$b = \cos θ + i \sin θ = \frac{1 - t^2}{1 + t^2} + \frac{2t}{1 + t^2}i$$

where $$t = \tan \frac θ2 ∈ ℝ$$. The $$x$$ coefficient is now

$$1 = a\overline b + \overline ab = \frac{1 - 4st - t^2}{1 + t^2},$$

so we must have

$$s = -\frac t2.$$

Finally, the $$x^2$$ coefficient is

$$0 = a\overline a + b + \overline b = \frac{t^4 - 6t^2 + 9}{4t^2 + 4} = \frac{(t^2 - 3)^2}{4t^2 + 4},$$

so $$t = \sqrt3$$ (or $$-\sqrt3$$, which would give the same solution with the quadratics swapped). Then $$s = -\frac{\sqrt3}2$$, $$a = \frac{1 - i\sqrt3}2$$, $$b = \frac{-1 + i\sqrt3}2$$, and the above factorization becomes

$$\left(x^2 + \frac{1 - i\sqrt3}2x + \frac{-1 + i\sqrt3}2\right)\left(x^2 + \frac{1 + i\sqrt3}2x + \frac{-1 - i\sqrt3}2\right).$$

– User
Oct 8, 2022 at 11:47

You can easily observe that, the expression $$(x-1)^2$$ is almost included in the polynomial $$P(x):=x^4+x^3-2x+1$$.

Let's rewrite your polynomial as follows:

\begin{align}P(x)=x^4+x^3-\color{red}{x^2}+\color{red}{x^2}-2x+1\end{align}

Based on this observation, we will represent the polynomial $$P(x)$$ as a "quadratic" polynomial:

\begin{align}P(x):&=x^4+x^3-x^2+(x-1)^2\\ &=x^2(x^2+x-1)+(x-1)^2\\ &=x^4+x^2(x-1)+(x-1)^2\\ &=\color{red}{(x-1)^2}+\color{blue}{x^2}\color{red}{(x-1)}+\color{blue}{x^4}\end{align}

Letting $$x-1=u$$ and writing $$P(x)=0$$ leads to:

\begin{align}&\color{red}{u^2}+\color{blue}{x^2}\color{red}{u}+\color{blue}{x^4}=0\\ &\Delta_{\color{red}{u}}=x^4-4x^4=-3x^4\\ \implies &u_{1,2}=\frac{-x^2\pm i\sqrt 3x^2}{2}\\ \implies &u_{1,2}=x^2\left(\frac{-1\pm i\sqrt 3}{2}\right)\\ \implies &x-1=x^2\left(\frac{-1\pm i\sqrt 3}{2}\right)\\ \implies &x^2\left(\frac{-1\pm i\sqrt 3}{2}\right)-x+1=0.\end{align}

Now, this is obvious that the given polynomial has no real roots.

Finally, you can complete the solution by using the quadratic formula to find all complex roots.

HINT.

$$4(x^4+x^3-2x+1)=(2x^2+x-1)^2+3(x-1)^2$$ $$=\big(2x^2+(1+i\sqrt3\,)(x-1)\big)\big(2x^2+(1-i\sqrt3)(x-1)\big),$$

DETAILS.

1. The first quadratic term can be chosen in the form of $$(2x^2+x+a)^2,$$ which provides the coefficients near $$x^4$$ and $$x^3.$$

2. The difference between the left part and the first quadratic term should be a full square.

$$(x^2+e^{\theta i}x+e^{\phi i})(x^2+e^{-\theta i}x+e^{-\phi i})=x^4+2\cos\theta x^3+(1+2\cos\phi)x^2+2\cos(\theta-\phi)x+1$$.

Then $$\cos\theta=\frac{1}{2}$$, $$\cos\phi=-\frac{1}{2}$$, $$\cos(\theta-\phi)=-1.$$

Then $$\theta$$ is $$\pm\frac{\pi}{3}$$ and $$\phi$$ is $$\pm\frac{2\pi}{3}$$.

Their difference is $$\pm\pi$$. Let $$\theta=\frac{\pi}{3}$$ and $$\phi=-\frac{2\pi}{3}$$

$$(x^2+e^{\frac{\pi}{3}i}x+e^{-\frac{2\pi}{3} i})(x^2+e^{-\frac{\pi}{3} i}x+e^{\frac{2\pi}{3} i})=0.$$

Then we solve by quadratic formula.