Solve x^3+10x^2+28x+24=0

$$x ^ { 3 } + 10 x ^ { 2 } + 28 x + 24 = 0$$

Solve for x

$$x=-2$$

$$x=-6$$

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.

$$x=-2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+10x^{2}+28x+24 by x+2 to get x^{2}+8x+12. Solve the equation where the result equals to 0.

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

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.

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

Do the calculations.

$$x=\frac{-8±4}{2}$$

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

$$x=-6 x=-2$$

List all found solutions.

$$x=-2 x=-6$$