Solve 10x^4+33x^3+31x^2+73x-210=0

$$10 x ^ { 4 } + 33 x ^ { 3 } + 31 x ^ { 2 } + 73 x - 210 = 0$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{19}i-1}{2}\approx -0.5-2.179449472i$$

$$x = \frac{6}{5} = 1\frac{1}{5} = 1.2$$

$$x=\frac{-1+\sqrt{19}i}{2}\approx -0.5+2.179449472i$$

$$x = -\frac{7}{2} = -3\frac{1}{2} = -3.5$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -210 and q divides the leading coefficient 10. List all candidates \frac{p}{q}.

$$±21,±42,±105,±210,±\frac{21}{2},±\frac{105}{2},±7,±14,±35,±70,±\frac{21}{5},±\frac{42}{5},±\frac{7}{2},±\frac{35}{2},±3,±6,±15,±30,±\frac{21}{10},±\frac{3}{2},±\frac{15}{2},±\frac{7}{5},±\frac{14}{5},±1,±2,±5,±10,±\frac{7}{10},±\frac{3}{5},±\frac{6}{5},±\frac{1}{2},±\frac{5}{2},±\frac{3}{10},±\frac{1}{5},±\frac{2}{5},±\frac{1}{10}$$

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.

$$x=\frac{6}{5}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 10x^{4}+33x^{3}+31x^{2}+73x-210 by 5\left(x-\frac{6}{5}\right)=5x-6 to get 2x^{3}+9x^{2}+17x+35. Solve the equation where the result equals to 0.

$$2x^{3}+9x^{2}+17x+35=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 35 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±\frac{35}{2},±35,±\frac{7}{2},±7,±\frac{5}{2},±5,±\frac{1}{2},±1$$

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.

$$x=-\frac{7}{2}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}+9x^{2}+17x+35 by 2\left(x+\frac{7}{2}\right)=2x+7 to get x^{2}+x+5. Solve the equation where the result equals to 0.

$$x^{2}+x+5=0$$

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, 1 for b, and 5 for c in the quadratic formula.

$$x=\frac{-1±\sqrt{1^{2}-4\times 1\times 5}}{2}$$

Do the calculations.

$$x=\frac{-1±\sqrt{-19}}{2}$$

Solve the equation x^{2}+x+5=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{19}i-1}{2} x=\frac{-1+\sqrt{19}i}{2}$$

List all found solutions.

$$x=\frac{6}{5} x=-\frac{7}{2} x=\frac{-\sqrt{19}i-1}{2} x=\frac{-1+\sqrt{19}i}{2}$$

Solve for x

$$x = -\frac{7}{2} = -3\frac{1}{2} = -3.5$$

$$x = \frac{6}{5} = 1\frac{1}{5} = 1.2$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -210 and q divides the leading coefficient 10. List all candidates \frac{p}{q}.

$$±21,±42,±105,±210,±\frac{21}{2},±\frac{105}{2},±7,±14,±35,±70,±\frac{21}{5},±\frac{42}{5},±\frac{7}{2},±\frac{35}{2},±3,±6,±15,±30,±\frac{21}{10},±\frac{3}{2},±\frac{15}{2},±\frac{7}{5},±\frac{14}{5},±1,±2,±5,±10,±\frac{7}{10},±\frac{3}{5},±\frac{6}{5},±\frac{1}{2},±\frac{5}{2},±\frac{3}{10},±\frac{1}{5},±\frac{2}{5},±\frac{1}{10}$$

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.

$$x=\frac{6}{5}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 10x^{4}+33x^{3}+31x^{2}+73x-210 by 5\left(x-\frac{6}{5}\right)=5x-6 to get 2x^{3}+9x^{2}+17x+35. Solve the equation where the result equals to 0.

$$2x^{3}+9x^{2}+17x+35=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 35 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±\frac{35}{2},±35,±\frac{7}{2},±7,±\frac{5}{2},±5,±\frac{1}{2},±1$$

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.

$$x=-\frac{7}{2}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}+9x^{2}+17x+35 by 2\left(x+\frac{7}{2}\right)=2x+7 to get x^{2}+x+5. Solve the equation where the result equals to 0.

$$x^{2}+x+5=0$$

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, 1 for b, and 5 for c in the quadratic formula.

$$x=\frac{-1±\sqrt{1^{2}-4\times 1\times 5}}{2}$$

Do the calculations.

$$x=\frac{-1±\sqrt{-19}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=\frac{6}{5} x=-\frac{7}{2}$$