Solve 24x^3-2x^2-5x+1=0

$$24 x ^ { 3 } - 2 x ^ { 2 } - 5 x + 1 = 0$$

Solve for x

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

$$x=\frac{1}{3}\approx 0.333333333$$

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

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 24. List all candidates \frac{p}{q}.

$$±\frac{1}{24},±\frac{1}{12},±\frac{1}{8},±\frac{1}{6},±\frac{1}{4},±\frac{1}{3},±\frac{1}{2},±1$$

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.

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

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 24x^{3}-2x^{2}-5x+1 by 4\left(x-\frac{1}{4}\right)=4x-1 to get 6x^{2}+x-1. Solve the equation where the result equals to 0.

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

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

$$x=\frac{-1±\sqrt{1^{2}-4\times 6\left(-1\right)}}{2\times 6}$$

Do the calculations.

$$x=\frac{-1±5}{12}$$

Solve the equation 6x^{2}+x-1=0 when ± is plus and when ± is minus.

$$x=-\frac{1}{2} x=\frac{1}{3}$$

List all found solutions.

$$x=\frac{1}{4} x=-\frac{1}{2} x=\frac{1}{3}$$