Solve 40x^3-133x^2+40x-3=0

$$40 x ^ { 3 } - 133 x ^ { 2 } + 40 x - 3 = 0$$

Solve for x

$$x=\frac{1}{5}=0.2$$

$$x=3$$

$$x=\frac{1}{8}=0.125$$

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 -3 and q divides the leading coefficient 40. List all candidates \frac{p}{q}.

$$±\frac{3}{40},±\frac{3}{20},±\frac{3}{10},±\frac{3}{8},±\frac{3}{5},±\frac{3}{4},±\frac{3}{2},±3,±\frac{1}{40},±\frac{1}{20},±\frac{1}{10},±\frac{1}{8},±\frac{1}{5},±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 40x^{3}-133x^{2}+40x-3 by x-3 to get 40x^{2}-13x+1. Solve the equation where the result equals to 0.

$$40x^{2}-13x+1=0$$

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 40 for a, -13 for b, and 1 for c in the quadratic formula.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 40\times 1}}{2\times 40}$$

Do the calculations.

$$x=\frac{13±3}{80}$$

Solve the equation 40x^{2}-13x+1=0 when ± is plus and when ± is minus.

$$x=\frac{1}{8} x=\frac{1}{5}$$

List all found solutions.

$$x=3 x=\frac{1}{8} x=\frac{1}{5}$$