Solve 2x^3+13x^2+13x-10=0

$$2 x ^ { 3 } + 13 x ^ { 2 } + 13 x - 10 = 0$$

Solve for x

$$x=-2$$

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

$$x=-5$$

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 -10 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±5,±10,±\frac{5}{2},±1,±2,±\frac{1}{2}$$

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.

$$x=-2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}+13x^{2}+13x-10 by x+2 to get 2x^{2}+9x-5. Solve the equation where the result equals to 0.

$$2x^{2}+9x-5=0$$

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, 9 for b, and -5 for c in the quadratic formula.

$$x=\frac{-9±\sqrt{9^{2}-4\times 2\left(-5\right)}}{2\times 2}$$

Do the calculations.

$$x=\frac{-9±11}{4}$$

Solve the equation 2x^{2}+9x-5=0 when ± is plus and when ± is minus.

$$x=-5 x=\frac{1}{2}$$

List all found solutions.

$$x=-2 x=-5 x=\frac{1}{2}$$