Solve the equation : $x^5-6x^4-17x^3+17x^2+6x-1=0$

By using Quadratic formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Derive the equation Given equation, $x^5-6x^4-17x^3+17x^2+6x-1=0$ Consider f(x) $=x^5-6x^4-17x^3+17x^2+6x-1$ Notice that this is a reciprocal equation of odd degree which has the opposite signs of the first and last term. $\therefore (x-1)$ is one factor of the given equation and the quotient is another reciprocal function which has same signs of the first and last term. $\therefore \mathrm{f}\left(\mathrm{x}\right)=(\mathrm{x}-1)\left({\mathrm{Ax}}^4+{\mathrm{Bx}}^3+{\mathrm{Cx}}^2+\mathrm{Bx}+\mathrm{A}\right)$ Comparing the coefficient, we have $\mathrm{A}=1,\mathrm{ }\mathrm{B}=-5,\mathrm{C}=-22$ $f\left(x\right)=(x-1)\left(x^4-5x^3-22x^2-5x+1\right)$ Consider $g\left(x\right)=x^4-5x^3-22x^2-5x+1=\left(x^4+1\right)-5\left(x^3+x\right)-22x^2$

Divide by ${\mathrm{x}}^2$ into the given equation We need to find the roots of $g\mathrm{\backslash mleft}\left(x\mathrm{\backslash mright}\right)=0$ $\left(x^2+x^{-2}\right)-5\left(x+x^{-1}\right)-22=0$

Substitute $x+x-1=y$ in the above equation $\left({\mathrm{y}}^2-2\right)-5\mathrm{y}-22=0⟹{\mathrm{y}}^2-5\mathrm{y}-24=0⟹(\mathrm{y}-8)(\mathrm{y}+3)=0$ $\therefore x+x^{-1}=8\text{ and }x+x^{-1}=3$ Solving the first quadratic equations we have, $\mathrm{x}=4\pm \sqrt{15}$ Solving the second quadratic equations we have, $\mathrm{x}=\frac{3\pm \sqrt{5}}{2}$

Roots of the given equation are $1,4\pm \sqrt{15},\frac{3\pm \sqrt{5}}{2}$