Solve x^3+6x^2+8x-192=0

$$x ^ { 3 } + 6 x ^ { 2 } + 8 x - 192 = 0$$

Solve for x (complex solution)

$$x=-\sqrt{23}i-5\approx -5-4.795831523i$$

$$x=4$$

$$x=-5+\sqrt{23}i\approx -5+4.795831523i$$

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 -192 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±192,±96,±64,±48,±32,±24,±16,±12,±8,±6,±4,±3,±2,±1$$

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.

$$x=4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+6x^{2}+8x-192 by x-4 to get x^{2}+10x+48. Solve the equation where the result equals to 0.

$$x^{2}+10x+48=0$$

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, 10 for b, and 48 for c in the quadratic formula.

$$x=\frac{-10±\sqrt{10^{2}-4\times 1\times 48}}{2}$$

Do the calculations.

$$x=\frac{-10±\sqrt{-92}}{2}$$

Solve the equation x^{2}+10x+48=0 when ± is plus and when ± is minus.

$$x=-\sqrt{23}i-5 x=-5+\sqrt{23}i$$

List all found solutions.

$$x=4 x=-\sqrt{23}i-5 x=-5+\sqrt{23}i$$

Solve for x

$$x=4$$

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 -192 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±192,±96,±64,±48,±32,±24,±16,±12,±8,±6,±4,±3,±2,±1$$

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.

$$x=4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+6x^{2}+8x-192 by x-4 to get x^{2}+10x+48. Solve the equation where the result equals to 0.

$$x^{2}+10x+48=0$$

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, 10 for b, and 48 for c in the quadratic formula.

$$x=\frac{-10±\sqrt{10^{2}-4\times 1\times 48}}{2}$$

Do the calculations.

$$x=\frac{-10±\sqrt{-92}}{2}$$

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.

$$x=4$$