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

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

Solve for x (complex solution)

$$x=\frac{1+2\sqrt{6}i}{5}\approx 0.2+0.979795897i$$

$$x=\frac{-2\sqrt{6}i+1}{5}\approx 0.2-0.979795897i$$

$$x=-1$$

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 5 and q divides the leading coefficient 5. List all candidates \frac{p}{q}.

$$±1,±5,±\frac{1}{5}$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 5x^{3}+3x^{2}+3x+5 by x+1 to get 5x^{2}-2x+5. Solve the equation where the result equals to 0.

$$5x^{2}-2x+5=0$$

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 5 for a, -2 for b, and 5 for c in the quadratic formula.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 5\times 5}}{2\times 5}$$

Do the calculations.

$$x=\frac{2±\sqrt{-96}}{10}$$

Solve the equation 5x^{2}-2x+5=0 when ± is plus and when ± is minus.

$$x=\frac{-2i\sqrt{6}+1}{5} x=\frac{1+2i\sqrt{6}}{5}$$

List all found solutions.

$$x=-1 x=\frac{-2i\sqrt{6}+1}{5} x=\frac{1+2i\sqrt{6}}{5}$$

Solve for x

$$x=-1$$

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 5 and q divides the leading coefficient 5. List all candidates \frac{p}{q}.

$$±1,±5,±\frac{1}{5}$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 5x^{3}+3x^{2}+3x+5 by x+1 to get 5x^{2}-2x+5. Solve the equation where the result equals to 0.

$$5x^{2}-2x+5=0$$

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 5 for a, -2 for b, and 5 for c in the quadratic formula.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 5\times 5}}{2\times 5}$$

Do the calculations.

$$x=\frac{2±\sqrt{-96}}{10}$$

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.

$$x=-1$$