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Solve for x (complex solution)
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±64,±32,±16,±8,±4,±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 64 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.
x=-4
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}-4x+16=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+64 by x+4 to get x^{2}-4x+16. Solve the equation where the result equals to 0.
x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\times 16}}{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, -4 for b, and 16 for c in the quadratic formula.
x=\frac{4±\sqrt{-48}}{2}
Do the calculations.
x=-2i\sqrt{3}+2 x=2+2i\sqrt{3}
Solve the equation x^{2}-4x+16=0 when ± is plus and when ± is minus.
x=-4 x=-2i\sqrt{3}+2 x=2+2i\sqrt{3}
List all found solutions.
±64,±32,±16,±8,±4,±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 64 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.
x=-4
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}-4x+16=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+64 by x+4 to get x^{2}-4x+16. Solve the equation where the result equals to 0.
x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\times 16}}{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, -4 for b, and 16 for c in the quadratic formula.
x=\frac{4±\sqrt{-48}}{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=-4
List all found solutions.