Solve 8x^3+125=0

$$8 x ^ { 3 } + 125 = 0$$

Solve for x (complex solution)

$$x=\frac{-5\sqrt{3}i+5}{4}\approx 1.25-2.165063509i$$

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

$$x=\frac{5+5\sqrt{3}i}{4}\approx 1.25+2.165063509i$$

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

$$±\frac{125}{8},±\frac{125}{4},±\frac{125}{2},±125,±\frac{25}{8},±\frac{25}{4},±\frac{25}{2},±25,±\frac{5}{8},±\frac{5}{4},±\frac{5}{2},±5,±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=-\frac{5}{2}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 8x^{3}+125 by 2\left(x+\frac{5}{2}\right)=2x+5 to get 4x^{2}-10x+25. Solve the equation where the result equals to 0.

$$4x^{2}-10x+25=0$$

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 4 for a, -10 for b, and 25 for c in the quadratic formula.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 4\times 25}}{2\times 4}$$

Do the calculations.

$$x=\frac{10±\sqrt{-300}}{8}$$

Solve the equation 4x^{2}-10x+25=0 when ± is plus and when ± is minus.

$$x=\frac{-5i\sqrt{3}+5}{4} x=\frac{5+5i\sqrt{3}}{4}$$

List all found solutions.

$$x=-\frac{5}{2} x=\frac{-5i\sqrt{3}+5}{4} x=\frac{5+5i\sqrt{3}}{4}$$

Solve for x

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

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

$$±\frac{125}{8},±\frac{125}{4},±\frac{125}{2},±125,±\frac{25}{8},±\frac{25}{4},±\frac{25}{2},±25,±\frac{5}{8},±\frac{5}{4},±\frac{5}{2},±5,±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=-\frac{5}{2}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 8x^{3}+125 by 2\left(x+\frac{5}{2}\right)=2x+5 to get 4x^{2}-10x+25. Solve the equation where the result equals to 0.

$$4x^{2}-10x+25=0$$

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 4 for a, -10 for b, and 25 for c in the quadratic formula.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 4\times 25}}{2\times 4}$$

Do the calculations.

$$x=\frac{10±\sqrt{-300}}{8}$$

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.

$$x=-\frac{5}{2}$$