Solve 2x^3+3x^2-32x+15=0

$$2 x ^ { 3 } + 3 x ^ { 2 } - 32 x + 15 = 0$$

Solve for x

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

$$x=3$$

$$x=-5$$

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

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

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}+3x^{2}-32x+15 by x-3 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.

$$x=3 x=-5 x=\frac{1}{2}$$