Solve (3x-1)(x^2-8)/x+1=0

$$\frac { ( 3 x - 1 ) ( x ^ { 2 } - 8 ) } { x + 1 } = 0$$

Solve for x

$$x=-2\sqrt{2}\approx -2.828427125$$

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

$$x=2\sqrt{2}\approx 2.828427125$$

Solution Steps

Variable x cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by x+1.

$$\left(3x-1\right)\left(x^{2}-8\right)=0$$

Use the distributive property to multiply 3x-1 by x^{2}-8.

$$3x^{3}-24x-x^{2}+8=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$3x^{3}-x^{2}-24x+8=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 8 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

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

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.

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

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

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

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 1 for a, 0 for b, and -8 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\left(-8\right)}}{2}$$

Do the calculations.

$$x=\frac{0±4\sqrt{2}}{2}$$

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

$$x=-2\sqrt{2} x=2\sqrt{2}$$

List all found solutions.

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