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Solve for x (complex solution)
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x^{3}+x^{2}-392=0
Subtract 392 from both sides.
±392,±196,±98,±56,±49,±28,±14,±8,±7,±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 -392 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.
x=7
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}+8x+56=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+x^{2}-392 by x-7 to get x^{2}+8x+56. Solve the equation where the result equals to 0.
x=\frac{-8±\sqrt{8^{2}-4\times 1\times 56}}{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, 8 for b, and 56 for c in the quadratic formula.
x=\frac{-8±\sqrt{-160}}{2}
Do the calculations.
x=-2i\sqrt{10}-4 x=-4+2i\sqrt{10}
Solve the equation x^{2}+8x+56=0 when ± is plus and when ± is minus.
x=7 x=-2i\sqrt{10}-4 x=-4+2i\sqrt{10}
List all found solutions.
x^{3}+x^{2}-392=0
Subtract 392 from both sides.
±392,±196,±98,±56,±49,±28,±14,±8,±7,±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 -392 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.
x=7
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}+8x+56=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+x^{2}-392 by x-7 to get x^{2}+8x+56. Solve the equation where the result equals to 0.
x=\frac{-8±\sqrt{8^{2}-4\times 1\times 56}}{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, 8 for b, and 56 for c in the quadratic formula.
x=\frac{-8±\sqrt{-160}}{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=7
List all found solutions.