Solve 5x^3-3x^2-32x-12=0

$$5 { x }^{ 3 } -3 { x }^{ 2 } -32x-12 = 0$$

Solve for x

$$x=-2$$

$$x=3$$

$$x=-\frac{2}{5}=-0.4$$

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 -12 and q divides the leading coefficient 5. List all candidates \frac{p}{q}.

$$±\frac{12}{5},±12,±\frac{6}{5},±6,±\frac{4}{5},±4,±\frac{3}{5},±3,±\frac{2}{5},±2,±\frac{1}{5},±1$$

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.

$$x=-2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 5x^{3}-3x^{2}-32x-12 by x+2 to get 5x^{2}-13x-6. Solve the equation where the result equals to 0.

$$5x^{2}-13x-6=0$$

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 5 for a, -13 for b, and -6 for c in the quadratic formula.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 5\left(-6\right)}}{2\times 5}$$

Do the calculations.

$$x=\frac{13±17}{10}$$

Solve the equation 5x^{2}-13x-6=0 when ± is plus and when ± is minus.

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

List all found solutions.

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