Solve x^3+5x^2+4x+20=0

$${ x }^{ 3 } +5 { x }^{ 2 } +4x+20 = 0$$

Solve for x

$$x=-5$$

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

$$±20,±10,±5,±4,±2,±1$$

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.

$$x=-5$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+5x^{2}+4x+20 by x+5 to get x^{2}+4. Solve the equation where the result equals to 0.

$$x^{2}+4=0$$

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, 0 for b, and 4 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\times 4}}{2}$$

Do the calculations.

$$x=\frac{0±\sqrt{-16}}{2}$$

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

$$x\in \emptyset$$

List all found solutions.

$$x=-5$$

Solve for x (complex solution)

$$x=2i$$

$$x=-2i$$

$$x=-5$$