Solve 2z^3+3z^2+8z-5=0

$$2 z ^ { 3 } + 3 z ^ { 2 } + 8 z - 5 = 0$$

Solve for z (complex solution)

$$z=-1-2i$$

$$z=\frac{1}{2}=0.5$$

$$z=-1+2i$$

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 -5 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±\frac{5}{2},±5,±\frac{1}{2},±1$$

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.

$$z=\frac{1}{2}$$

By Factor theorem, z-k is a factor of the polynomial for each root k. Divide 2z^{3}+3z^{2}+8z-5 by 2\left(z-\frac{1}{2}\right)=2z-1 to get z^{2}+2z+5. Solve the equation where the result equals to 0.

$$z^{2}+2z+5=0$$

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

$$z=\frac{-2±\sqrt{2^{2}-4\times 1\times 5}}{2}$$

Do the calculations.

$$z=\frac{-2±\sqrt{-16}}{2}$$

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

$$z=-1-2i z=-1+2i$$

List all found solutions.

$$z=\frac{1}{2} z=-1-2i z=-1+2i$$

Solve for z

$$z=\frac{1}{2}=0.5$$

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 -5 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±\frac{5}{2},±5,±\frac{1}{2},±1$$

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.

$$z=\frac{1}{2}$$

By Factor theorem, z-k is a factor of the polynomial for each root k. Divide 2z^{3}+3z^{2}+8z-5 by 2\left(z-\frac{1}{2}\right)=2z-1 to get z^{2}+2z+5. Solve the equation where the result equals to 0.

$$z^{2}+2z+5=0$$

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

$$z=\frac{-2±\sqrt{2^{2}-4\times 1\times 5}}{2}$$

Do the calculations.

$$z=\frac{-2±\sqrt{-16}}{2}$$

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

$$z\in \emptyset$$

List all found solutions.

$$z=\frac{1}{2}$$