Solve 2x^3-x^2-5x-2=0

$$2 x ^ { 3 } - x ^ { 2 } - 5 x - 2 = 0$$

Solve for x

$$x=-1$$

$$x=2$$

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

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 -2 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±1,±2,±\frac{1}{2}$$

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 2x^{3}-x^{2}-5x-2 by x+1 to get 2x^{2}-3x-2. Solve the equation where the result equals to 0.

$$2x^{2}-3x-2=0$$

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 2 for a, -3 for b, and -2 for c in the quadratic formula.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\left(-2\right)}}{2\times 2}$$

Do the calculations.

$$x=\frac{3±5}{4}$$

Solve the equation 2x^{2}-3x-2=0 when ± is plus and when ± is minus.

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

List all found solutions.

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