Solve 6x^5+13x^4-18x^3-37x^2+16x+20=0

$$6 { x }^{ 5 } +13 { x }^{ 4 } -18 { x }^{ 3 } -37 { x }^{ 2 } +16x+20 = 0$$

Solve for x

$$x = \frac{\sqrt{41} - 1}{4} \approx 1.350781059$$

$$x=\frac{-\sqrt{41}-1}{4}\approx -1.850781059$$

$$x=1$$

$$x=-2$$

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

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 20 and q divides the leading coefficient 6. List all candidates \frac{p}{q}.

$$±\frac{10}{3},±\frac{20}{3},±10,±20,±\frac{5}{3},±5,±\frac{5}{6},±\frac{5}{2},±\frac{2}{3},±\frac{4}{3},±2,±4,±\frac{1}{3},±1,±\frac{1}{6},±\frac{1}{2}$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 6x^{5}+13x^{4}-18x^{3}-37x^{2}+16x+20 by x-1 to get 6x^{4}+19x^{3}+x^{2}-36x-20. Solve the equation where the result equals to 0.

$$6x^{4}+19x^{3}+x^{2}-36x-20=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -20 and q divides the leading coefficient 6. List all candidates \frac{p}{q}.

$$±\frac{10}{3},±\frac{20}{3},±10,±20,±\frac{5}{3},±5,±\frac{5}{6},±\frac{5}{2},±\frac{2}{3},±\frac{4}{3},±2,±4,±\frac{1}{3},±1,±\frac{1}{6},±\frac{1}{2}$$

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.

$$x=-2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 6x^{4}+19x^{3}+x^{2}-36x-20 by x+2 to get 6x^{3}+7x^{2}-13x-10. Solve the equation where the result equals to 0.

$$6x^{3}+7x^{2}-13x-10=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -10 and q divides the leading coefficient 6. List all candidates \frac{p}{q}.

$$±\frac{5}{3},±\frac{10}{3},±5,±10,±\frac{5}{6},±\frac{5}{2},±\frac{1}{3},±\frac{2}{3},±1,±2,±\frac{1}{6},±\frac{1}{2}$$

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 6x^{3}+7x^{2}-13x-10 by 3\left(x+\frac{2}{3}\right)=3x+2 to get 2x^{2}+x-5. Solve the equation where the result equals to 0.

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

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

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

Do the calculations.

$$x=\frac{-1±\sqrt{41}}{4}$$

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

$$x=\frac{-\sqrt{41}-1}{4} x=\frac{\sqrt{41}-1}{4}$$

List all found solutions.

$$x=1 x=-2 x=-\frac{2}{3} x=\frac{-\sqrt{41}-1}{4} x=\frac{\sqrt{41}-1}{4}$$