Solve 2x^3+3x^2-17x-30=0

$$2 x ^ { 3 } + 3 x ^ { 2 } - 17 x - 30 = 0$$

Solve for x

$$x=-2$$

$$x=3$$

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

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 -30 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±15,±30,±\frac{15}{2},±5,±10,±3,±6,±\frac{5}{2},±\frac{3}{2},±1,±2,±\frac{1}{2}$$

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}+3x^{2}-17x-30 by x+2 to get 2x^{2}-x-15. Solve the equation where the result equals to 0.

$$2x^{2}-x-15=0$$

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, -1 for b, and -15 for c in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 2\left(-15\right)}}{2\times 2}$$

Do the calculations.

$$x=\frac{1±11}{4}$$

Solve the equation 2x^{2}-x-15=0 when ± is plus and when ± is minus.

$$x=-\frac{5}{2} x=3$$

List all found solutions.

$$x=-2 x=-\frac{5}{2} x=3$$