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Solve for x (complex solution)
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27x^{3}+162x^{2}+324x+280=0
Expand the expression.
±\frac{280}{27},±\frac{280}{9},±\frac{280}{3},±280,±\frac{140}{27},±\frac{140}{9},±\frac{140}{3},±140,±\frac{70}{27},±\frac{70}{9},±\frac{70}{3},±70,±\frac{56}{27},±\frac{56}{9},±\frac{56}{3},±56,±\frac{40}{27},±\frac{40}{9},±\frac{40}{3},±40,±\frac{35}{27},±\frac{35}{9},±\frac{35}{3},±35,±\frac{28}{27},±\frac{28}{9},±\frac{28}{3},±28,±\frac{20}{27},±\frac{20}{9},±\frac{20}{3},±20,±\frac{14}{27},±\frac{14}{9},±\frac{14}{3},±14,±\frac{10}{27},±\frac{10}{9},±\frac{10}{3},±10,±\frac{8}{27},±\frac{8}{9},±\frac{8}{3},±8,±\frac{7}{27},±\frac{7}{9},±\frac{7}{3},±7,±\frac{5}{27},±\frac{5}{9},±\frac{5}{3},±5,±\frac{4}{27},±\frac{4}{9},±\frac{4}{3},±4,±\frac{2}{27},±\frac{2}{9},±\frac{2}{3},±2,±\frac{1}{27},±\frac{1}{9},±\frac{1}{3},±1
By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 280 and q divides the leading coefficient 27. List all candidates \frac{p}{q}.
x=-\frac{10}{3}
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.
9x^{2}+24x+28=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 27x^{3}+162x^{2}+324x+280 by 3\left(x+\frac{10}{3}\right)=3x+10 to get 9x^{2}+24x+28. Solve the equation where the result equals to 0.
x=\frac{-24±\sqrt{24^{2}-4\times 9\times 28}}{2\times 9}
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 9 for a, 24 for b, and 28 for c in the quadratic formula.
x=\frac{-24±\sqrt{-432}}{18}
Do the calculations.
x=\frac{-2i\sqrt{3}-4}{3} x=\frac{-4+2i\sqrt{3}}{3}
Solve the equation 9x^{2}+24x+28=0 when ± is plus and when ± is minus.
x=-\frac{10}{3} x=\frac{-2i\sqrt{3}-4}{3} x=\frac{-4+2i\sqrt{3}}{3}
List all found solutions.
27x^{3}+162x^{2}+324x+280=0
Expand the expression.
±\frac{280}{27},±\frac{280}{9},±\frac{280}{3},±280,±\frac{140}{27},±\frac{140}{9},±\frac{140}{3},±140,±\frac{70}{27},±\frac{70}{9},±\frac{70}{3},±70,±\frac{56}{27},±\frac{56}{9},±\frac{56}{3},±56,±\frac{40}{27},±\frac{40}{9},±\frac{40}{3},±40,±\frac{35}{27},±\frac{35}{9},±\frac{35}{3},±35,±\frac{28}{27},±\frac{28}{9},±\frac{28}{3},±28,±\frac{20}{27},±\frac{20}{9},±\frac{20}{3},±20,±\frac{14}{27},±\frac{14}{9},±\frac{14}{3},±14,±\frac{10}{27},±\frac{10}{9},±\frac{10}{3},±10,±\frac{8}{27},±\frac{8}{9},±\frac{8}{3},±8,±\frac{7}{27},±\frac{7}{9},±\frac{7}{3},±7,±\frac{5}{27},±\frac{5}{9},±\frac{5}{3},±5,±\frac{4}{27},±\frac{4}{9},±\frac{4}{3},±4,±\frac{2}{27},±\frac{2}{9},±\frac{2}{3},±2,±\frac{1}{27},±\frac{1}{9},±\frac{1}{3},±1
By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 280 and q divides the leading coefficient 27. List all candidates \frac{p}{q}.
x=-\frac{10}{3}
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.
9x^{2}+24x+28=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 27x^{3}+162x^{2}+324x+280 by 3\left(x+\frac{10}{3}\right)=3x+10 to get 9x^{2}+24x+28. Solve the equation where the result equals to 0.
x=\frac{-24±\sqrt{24^{2}-4\times 9\times 28}}{2\times 9}
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 9 for a, 24 for b, and 28 for c in the quadratic formula.
x=\frac{-24±\sqrt{-432}}{18}
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=-\frac{10}{3}
List all found solutions.