Solve 2x^3+6x^2+7x+60=0

$$2 { x }^{ 3 } +6 { x }^{ 2 } +7x+60 = 0$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{29}i+1}{2}\approx 0.5-2.692582404i$$

$$x=-4$$

$$x=\frac{1+\sqrt{29}i}{2}\approx 0.5+2.692582404i$$

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

$$±30,±60,±15,±10,±20,±\frac{15}{2},±6,±12,±5,±3,±\frac{5}{2},±2,±4,±\frac{3}{2},±1,±\frac{1}{2}$$

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.

$$x=-4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}+6x^{2}+7x+60 by x+4 to get 2x^{2}-2x+15. Solve the equation where the result equals to 0.

$$2x^{2}-2x+15=0$$

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, -2 for b, and 15 for c in the quadratic formula.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 2\times 15}}{2\times 2}$$

Do the calculations.

$$x=\frac{2±\sqrt{-116}}{4}$$

Solve the equation 2x^{2}-2x+15=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{29}i+1}{2} x=\frac{1+\sqrt{29}i}{2}$$

List all found solutions.

$$x=-4 x=\frac{-\sqrt{29}i+1}{2} x=\frac{1+\sqrt{29}i}{2}$$

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

$$±30,±60,±15,±10,±20,±\frac{15}{2},±6,±12,±5,±3,±\frac{5}{2},±2,±4,±\frac{3}{2},±1,±\frac{1}{2}$$

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.

$$x=-4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}+6x^{2}+7x+60 by x+4 to get 2x^{2}-2x+15. Solve the equation where the result equals to 0.

$$2x^{2}-2x+15=0$$

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, -2 for b, and 15 for c in the quadratic formula.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 2\times 15}}{2\times 2}$$

Do the calculations.

$$x=\frac{2±\sqrt{-116}}{4}$$

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.

$$x=-4$$