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Solve for x (complex solution)
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±90,±45,±30,±18,±15,±10,±9,±6,±5,±3,±2,±1
By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -90 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.
x=6
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}+6x+15=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-21x-90 by x-6 to get x^{2}+6x+15. Solve the equation where the result equals to 0.
x=\frac{-6±\sqrt{6^{2}-4\times 1\times 15}}{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 1 for a, 6 for b, and 15 for c in the quadratic formula.
x=\frac{-6±\sqrt{-24}}{2}
Do the calculations.
x=-\sqrt{6}i-3 x=-3+\sqrt{6}i
Solve the equation x^{2}+6x+15=0 when ± is plus and when ± is minus.
x=6 x=-\sqrt{6}i-3 x=-3+\sqrt{6}i
List all found solutions.
±90,±45,±30,±18,±15,±10,±9,±6,±5,±3,±2,±1
By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -90 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.
x=6
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}+6x+15=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-21x-90 by x-6 to get x^{2}+6x+15. Solve the equation where the result equals to 0.
x=\frac{-6±\sqrt{6^{2}-4\times 1\times 15}}{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 1 for a, 6 for b, and 15 for c in the quadratic formula.
x=\frac{-6±\sqrt{-24}}{2}
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=6
List all found solutions.