Solve 90x^4-9x^3-562x^2-29x+210=0

$$90 { x }^{ 4 } -9 { x }^{ 3 } -562 { x }^{ 2 } -29x+210 = 0$$

Solve for x

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

$$x=-\frac{2}{3}\approx -0.666666667$$

$$x=\frac{3}{5}=0.6$$

$$x = -\frac{7}{3} = -2\frac{1}{3} \approx -2.333333333$$

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 90. List all candidates \frac{p}{q}.

$$±\frac{7}{3},±\frac{14}{3},±7,±\frac{35}{3},±14,±21,±\frac{70}{3},±35,±42,±70,±105,±210,±\frac{7}{6},±\frac{7}{2},±\frac{35}{6},±\frac{21}{2},±\frac{35}{2},±\frac{105}{2},±\frac{7}{9},±\frac{14}{9},±\frac{35}{9},±\frac{70}{9},±\frac{7}{15},±\frac{14}{15},±\frac{7}{5},±\frac{14}{5},±\frac{21}{5},±\frac{42}{5},±\frac{7}{18},±\frac{35}{18},±\frac{1}{3},±\frac{2}{3},±1,±\frac{5}{3},±2,±3,±\frac{10}{3},±5,±6,±10,±15,±30,±\frac{7}{30},±\frac{7}{10},±\frac{21}{10},±\frac{1}{6},±\frac{1}{2},±\frac{5}{6},±\frac{3}{2},±\frac{5}{2},±\frac{15}{2},±\frac{7}{45},±\frac{14}{45},±\frac{1}{9},±\frac{2}{9},±\frac{5}{9},±\frac{10}{9},±\frac{7}{90},±\frac{1}{15},±\frac{2}{15},±\frac{1}{5},±\frac{2}{5},±\frac{3}{5},±\frac{6}{5},±\frac{1}{18},±\frac{5}{18},±\frac{1}{30},±\frac{1}{10},±\frac{3}{10},±\frac{1}{45},±\frac{2}{45},±\frac{1}{90}$$

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.

$$x=\frac{3}{5}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 90x^{4}-9x^{3}-562x^{2}-29x+210 by 5\left(x-\frac{3}{5}\right)=5x-3 to get 18x^{3}+9x^{2}-107x-70. Solve the equation where the result equals to 0.

$$18x^{3}+9x^{2}-107x-70=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -70 and q divides the leading coefficient 18. List all candidates \frac{p}{q}.

$$±\frac{35}{9},±\frac{70}{9},±\frac{35}{3},±\frac{70}{3},±35,±70,±\frac{35}{18},±\frac{35}{6},±\frac{35}{2},±\frac{7}{9},±\frac{14}{9},±\frac{7}{3},±\frac{14}{3},±7,±14,±\frac{5}{9},±\frac{10}{9},±\frac{5}{3},±\frac{10}{3},±5,±10,±\frac{7}{18},±\frac{7}{6},±\frac{7}{2},±\frac{5}{18},±\frac{5}{6},±\frac{5}{2},±\frac{1}{9},±\frac{2}{9},±\frac{1}{3},±\frac{2}{3},±1,±2,±\frac{1}{18},±\frac{1}{6},±\frac{1}{2}$$

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.

$$x=-\frac{2}{3}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 18x^{3}+9x^{2}-107x-70 by 3\left(x+\frac{2}{3}\right)=3x+2 to get 6x^{2}-x-35. Solve the equation where the result equals to 0.

$$6x^{2}-x-35=0$$

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 6 for a, -1 for b, and -35 for c in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 6\left(-35\right)}}{2\times 6}$$

Do the calculations.

$$x=\frac{1±29}{12}$$

Solve the equation 6x^{2}-x-35=0 when ± is plus and when ± is minus.

$$x=-\frac{7}{3} x=\frac{5}{2}$$

List all found solutions.

$$x=\frac{3}{5} x=-\frac{2}{3} x=-\frac{7}{3} x=\frac{5}{2}$$