Solve 2x^3+3x^2+4=0

$$2 x ^ { 3 } + 3 x ^ { 2 } + 4 = 0$$

Solve for x (complex solution)

$$x=\frac{1+\sqrt{15}i}{4}\approx 0.25+0.968245837i$$

$$x=\frac{-\sqrt{15}i+1}{4}\approx 0.25-0.968245837i$$

$$x=-2$$

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

$$±2,±4,±1,±\frac{1}{2}$$

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 2x^{3}+3x^{2}+4 by x+2 to get 2x^{2}-x+2. Solve the equation where the result equals to 0.

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

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

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 2\times 2}}{2\times 2}$$

Do the calculations.

$$x=\frac{1±\sqrt{-15}}{4}$$

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

$$x=\frac{-\sqrt{15}i+1}{4} x=\frac{1+\sqrt{15}i}{4}$$

List all found solutions.

$$x=-2 x=\frac{-\sqrt{15}i+1}{4} x=\frac{1+\sqrt{15}i}{4}$$

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

$$±2,±4,±1,±\frac{1}{2}$$

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 2x^{3}+3x^{2}+4 by x+2 to get 2x^{2}-x+2. Solve the equation where the result equals to 0.

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

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

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 2\times 2}}{2\times 2}$$

Do the calculations.

$$x=\frac{1±\sqrt{-15}}{4}$$

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.

$$x=-2$$