How can I solve this fourth-order polynomial equation?

Question

How can I solve this problem to get the value of $x$? $$x^4 - 5x+ 4=0$$

Answer 1

Following @GitGud 's idea and using rational root theorem, I found out that $x=1$ is a root of the equation. Hence, $$\left(x-1\right)\left(x^3+x^2+x-4\right)=0$$ So one solution is $x=1$ and the cubic factor does not have a pretty solution.

Edit:

The plot of the function shows only $2$ real roots. The second one is also close to $1$.

Answer 2

As shown in other answers, the rational roots theorem reduces the problem down to a cubic factoring:

$$x^4-5x+4=(x−1)(x^3+x^2+x−4)$$

To factor the cubic, start with the substitution $x=\frac{2\sqrt2}3y-\frac13$. You should end up with

$$\frac{16\sqrt2}{27}y^3+\frac{4\sqrt2}9y-\frac{115}{27}=0$$

Multiply by $\frac{27}{4\sqrt2}$ on both sides to get

$$4y^3+3y-\frac{115}{4\sqrt2}=0$$

$$4y^3+3y=\frac{115}{4\sqrt2}$$

Recall a hyperbolic trig identity:

$$\sinh(3\operatorname{arcsinh}(y))=4y^3+3y=\frac{115}{4\sqrt2}$$

$$y=\sinh\left(\frac13\operatorname{arcsinh}\left(\frac{115}{4\sqrt2}\right)\right)$$

$$x=\frac{2{\sqrt2}}3y-\frac13=\frac{2{\sqrt2}}3\sinh\left(\frac13\operatorname{arcsinh}\left(\frac{115}{4\sqrt2}\right)\right)-\frac13\approx1.15091108434$$

That is the second root. After factoring it out, the two remaining complex roots may be found via quadratic formula.

Alternatively, if one is able to handle complex exponents, the following form is more compact:

$$x=\frac{2{\sqrt2}}3\sinh\left(\frac{2k\pi i}3+\frac13\operatorname{arcsinh}\left(\frac{115}{4\sqrt2}\right)\right)-\frac13\qquad k=0,1,2$$