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Solve for x (complex solution)
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Solve for x
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4x^{3}+2x^{2}+x+7-49=0
Subtract 49 from both sides.
4x^{3}+2x^{2}+x-42=0
Subtract 49 from 7 to get -42.
±\frac{21}{2},±21,±42,±\frac{21}{4},±\frac{7}{2},±7,±14,±\frac{7}{4},±\frac{3}{2},±3,±6,±\frac{3}{4},±\frac{1}{2},±1,±2,±\frac{1}{4}
By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -42 and q divides the leading coefficient 4. 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.
4x^{2}+10x+21=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{3}+2x^{2}+x-42 by x-2 to get 4x^{2}+10x+21. Solve the equation where the result equals to 0.
x=\frac{-10±\sqrt{10^{2}-4\times 4\times 21}}{2\times 4}
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 4 for a, 10 for b, and 21 for c in the quadratic formula.
x=\frac{-10±\sqrt{-236}}{8}
Do the calculations.
x=\frac{-\sqrt{59}i-5}{4} x=\frac{-5+\sqrt{59}i}{4}
Solve the equation 4x^{2}+10x+21=0 when ± is plus and when ± is minus.
x=2 x=\frac{-\sqrt{59}i-5}{4} x=\frac{-5+\sqrt{59}i}{4}
List all found solutions.
4x^{3}+2x^{2}+x+7-49=0
Subtract 49 from both sides.
4x^{3}+2x^{2}+x-42=0
Subtract 49 from 7 to get -42.
±\frac{21}{2},±21,±42,±\frac{21}{4},±\frac{7}{2},±7,±14,±\frac{7}{4},±\frac{3}{2},±3,±6,±\frac{3}{4},±\frac{1}{2},±1,±2,±\frac{1}{4}
By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -42 and q divides the leading coefficient 4. 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.
4x^{2}+10x+21=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{3}+2x^{2}+x-42 by x-2 to get 4x^{2}+10x+21. Solve the equation where the result equals to 0.
x=\frac{-10±\sqrt{10^{2}-4\times 4\times 21}}{2\times 4}
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 4 for a, 10 for b, and 21 for c in the quadratic formula.
x=\frac{-10±\sqrt{-236}}{8}
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.