Solve u^3+6u^2-4u-24=0

$$u ^ { 3 } + 6 u ^ { 2 } - 4 u - 24 = 0$$

Solve for u

$$u=-2$$

$$u=2$$

$$u=-6$$

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 -24 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±24,±12,±8,±6,±4,±3,±2,±1$$

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.

$$u=2$$

By Factor theorem, u-k is a factor of the polynomial for each root k. Divide u^{3}+6u^{2}-4u-24 by u-2 to get u^{2}+8u+12. Solve the equation where the result equals to 0.

$$u^{2}+8u+12=0$$

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 1 for a, 8 for b, and 12 for c in the quadratic formula.

$$u=\frac{-8±\sqrt{8^{2}-4\times 1\times 12}}{2}$$

Do the calculations.

$$u=\frac{-8±4}{2}$$

Solve the equation u^{2}+8u+12=0 when ± is plus and when ± is minus.

$$u=-6 u=-2$$

List all found solutions.

$$u=2 u=-6 u=-2$$