Solve 41x^3+59x^2-22x-40=0

$$41 { x }^{ 3 } +59 { x }^{ 2 } -22x-40=0$$

Solve for x

$$x=\frac{\sqrt{1721}-9}{41}\approx 0.792315537$$

$$x=\frac{-\sqrt{1721}-9}{41}\approx -1.231339927$$

$$x=-1$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -40 and q divides the leading coefficient 41. List all candidates \frac{p}{q}.

$$±\frac{40}{41},±40,±\frac{20}{41},±20,±\frac{10}{41},±10,±\frac{8}{41},±8,±\frac{5}{41},±5,±\frac{4}{41},±4,±\frac{2}{41},±2,±\frac{1}{41},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 41x^{3}+59x^{2}-22x-40 by x+1 to get 41x^{2}+18x-40. Solve the equation where the result equals to 0.

$$41x^{2}+18x-40=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 41 for a, 18 for b, and -40 for c in the quadratic formula.

$$x=\frac{-18±\sqrt{18^{2}-4\times 41\left(-40\right)}}{2\times 41}$$

Do the calculations.

$$x=\frac{-18±2\sqrt{1721}}{82}$$

Solve the equation 41x^{2}+18x-40=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{1721}-9}{41} x=\frac{\sqrt{1721}-9}{41}$$

List all found solutions.

$$x=-1 x=\frac{-\sqrt{1721}-9}{41} x=\frac{\sqrt{1721}-9}{41}$$