Solve s^3+s^2+2s+8=0

$$s ^ { 3 } + s ^ { 2 } + 2 s + 8 = 0$$

Solve for s

$$s=-2$$

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

$$±8,±4,±2,±1$$

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.

$$s=-2$$

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

$$s^{2}-s+4=0$$

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, -1 for b, and 4 for c in the quadratic formula.

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

Do the calculations.

$$s=\frac{1±\sqrt{-15}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$s\in \emptyset$$

List all found solutions.

$$s=-2$$