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Solve for x (complex solution)
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±1,±2,±\frac{1}{2}
By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -2 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.
x=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.
2x^{2}+x+1=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}-3x^{2}-x-2 by x-2 to get 2x^{2}+x+1. Solve the equation where the result equals to 0.
x=\frac{-1±\sqrt{1^{2}-4\times 2\times 1}}{2\times 2}
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 1 for c in the quadratic formula.
x=\frac{-1±\sqrt{-7}}{4}
Do the calculations.
x=\frac{-\sqrt{7}i-1}{4} x=\frac{-1+\sqrt{7}i}{4}
Solve the equation 2x^{2}+x+1=0 when ± is plus and when ± is minus.
x=2 x=\frac{-\sqrt{7}i-1}{4} x=\frac{-1+\sqrt{7}i}{4}
List all found solutions.
±1,±2,±\frac{1}{2}
By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -2 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.
x=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.
2x^{2}+x+1=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}-3x^{2}-x-2 by x-2 to get 2x^{2}+x+1. Solve the equation where the result equals to 0.
x=\frac{-1±\sqrt{1^{2}-4\times 2\times 1}}{2\times 2}
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 1 for c in the quadratic formula.
x=\frac{-1±\sqrt{-7}}{4}
Do the calculations.
x\in \emptyset
Since the square root of a negative number is not defined in the real field, there are no solutions.
x=2
List all found solutions.