solve the equation $x^4-10x^3+26x^2-10x+1$

Question

Solve the equation

$$x^4-10x^3+26x^2-10x+1=0$$

I tried for the values of $x = \pm 1, \pm 2, \pm 3$ but I didn't get the solution.

Answer 1

If we divide $x^2$ to both side (since $x=0$ isn't a solution), we get $$x^2 - 10x + 26 - 10\dfrac{1}{x} + \dfrac{1}{x^2} = 0,$$ then the coefficients are "symmetric."

So if we let $A = x+\dfrac{1}{x}$, then $x^2 + \dfrac{1}{x^2} = A^2-2.$ Now the original equation can be written as a quadratic polynomial of $A$.

Answer 2

First of all, it was unnecessary to try $\pm 2, \pm 3$, see this:

https://www.britannica.com/science/rational-root-theorem

The thing you need to observe is that the polynomial is symmetrical. In such cases, try introducing the new variable $y=x+\frac{1}{x}$, to obtain a quadratic equation in $y$. (And then ultimately, quadratic equations in $x$.)

Answer 3

As you eliminated $\pm1$, there is no rational solution (rational root theorem). The symmetry of the polynomial suggests comparing it with $(x-\frac52)^4=x^4-10x^3+\frac{75}{2}x^2-10x+1$ and recall that $a^2-b^2=(a+b)(a-b)$.

Answer 4

This type of equation is called reciprocal equation. Note the symmetry of coefficient.

Since $x=0$ is not the solution of equation, you can divide the equation by $x^2$:

$$ x^2 -10x +26 -10/x + 1/x^2 =0 $$ $$( x^2 + 1/x^2) -10(x+1/x) +26=0$$

Now, let $y=x +1/x$ Then $y^2=(x +1/x)^2$ and therefore $y^2 -2=x^2 +1/x^2$

$$y^2 - 2 - 10y +26 =0$$ $$y^2 - 10y +24 =0$$ $$(y-6)(y-4)=0$$ $y=6$ or $y =4$

From here, you solve $x +1/x=6$ and $x +1/x=4$.