Solve x^3+6x^2+16x+16=0

$${ x }^{ 3 } +6 { x }^{ 2 } +16x+16 = 0$$

Solve for x (complex solution)

$$x=-2-2i$$

$$x=-2$$

$$x=-2+2i$$

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

$$±16,±8,±4,±2,±1$$

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.

$$x=-2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+6x^{2}+16x+16 by x+2 to get x^{2}+4x+8. Solve the equation where the result equals to 0.

$$x^{2}+4x+8=0$$

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, 4 for b, and 8 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{-4±\sqrt{-16}}{2}$$

Solve the equation x^{2}+4x+8=0 when ± is plus and when ± is minus.

$$x=-2-2i x=-2+2i$$

List all found solutions.

$$x=-2 x=-2-2i x=-2+2i$$

Solve for x

$$x=-2$$

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

$$±16,±8,±4,±2,±1$$

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.

$$x=-2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+6x^{2}+16x+16 by x+2 to get x^{2}+4x+8. Solve the equation where the result equals to 0.

$$x^{2}+4x+8=0$$

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, 4 for b, and 8 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{-4±\sqrt{-16}}{2}$$

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.

$$x=-2$$