Solve 64x^3-24x^2-6x+1=0

$$64 x ^ { 3 } - 24 x ^ { 2 } - 6 x + 1 = 0$$

Solve for x

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

$$x=\frac{1}{2}=0.5$$

$$x=-\frac{1}{4}=-0.25$$

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 1 and q divides the leading coefficient 64. List all candidates \frac{p}{q}.

$$±\frac{1}{64},±\frac{1}{32},±\frac{1}{16},±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

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.

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

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 64x^{3}-24x^{2}-6x+1 by 8\left(x-\frac{1}{8}\right)=8x-1 to get 8x^{2}-2x-1. Solve the equation where the result equals to 0.

$$8x^{2}-2x-1=0$$

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 8 for a, -2 for b, and -1 for c in the quadratic formula.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 8\left(-1\right)}}{2\times 8}$$

Do the calculations.

$$x=\frac{2±6}{16}$$

Solve the equation 8x^{2}-2x-1=0 when ± is plus and when ± is minus.

$$x=-\frac{1}{4} x=\frac{1}{2}$$

List all found solutions.

$$x=\frac{1}{8} x=-\frac{1}{4} x=\frac{1}{2}$$