Solve 2x^4-11x^3+26x^2-66x+84=0

$$2 x ^ { 4 } - 11 x ^ { 3 } + 26 x ^ { 2 } - 66 x + 84 = 0$$

Solve for x (complex solution)

$$x=2$$

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

$$x=\sqrt{6}i\approx 2.449489743i$$

$$x=-\sqrt{6}i\approx -0-2.449489743i$$

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 84 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±42,±84,±21,±14,±28,±\frac{21}{2},±7,±6,±12,±\frac{7}{2},±3,±2,±4,±\frac{3}{2},±1,±\frac{1}{2}$$

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 2x^{4}-11x^{3}+26x^{2}-66x+84 by x-2 to get 2x^{3}-7x^{2}+12x-42. Solve the equation where the result equals to 0.

$$2x^{3}-7x^{2}+12x-42=0$$

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

$$±21,±42,±\frac{21}{2},±7,±14,±\frac{7}{2},±3,±6,±\frac{3}{2},±1,±2,±\frac{1}{2}$$

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.

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

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}-7x^{2}+12x-42 by 2\left(x-\frac{7}{2}\right)=2x-7 to get x^{2}+6. Solve the equation where the result equals to 0.

$$x^{2}+6=0$$

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, 0 for b, and 6 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\times 6}}{2}$$

Do the calculations.

$$x=\frac{0±\sqrt{-24}}{2}$$

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

$$x=-\sqrt{6}i x=\sqrt{6}i$$

List all found solutions.

$$x=2 x=\frac{7}{2} x=-\sqrt{6}i x=\sqrt{6}i$$

Solve for x

$$x=2$$

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

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 84 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±42,±84,±21,±14,±28,±\frac{21}{2},±7,±6,±12,±\frac{7}{2},±3,±2,±4,±\frac{3}{2},±1,±\frac{1}{2}$$

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 2x^{4}-11x^{3}+26x^{2}-66x+84 by x-2 to get 2x^{3}-7x^{2}+12x-42. Solve the equation where the result equals to 0.

$$2x^{3}-7x^{2}+12x-42=0$$

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

$$±21,±42,±\frac{21}{2},±7,±14,±\frac{7}{2},±3,±6,±\frac{3}{2},±1,±2,±\frac{1}{2}$$

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.

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

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}-7x^{2}+12x-42 by 2\left(x-\frac{7}{2}\right)=2x-7 to get x^{2}+6. Solve the equation where the result equals to 0.

$$x^{2}+6=0$$

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, 0 for b, and 6 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\times 6}}{2}$$

Do the calculations.

$$x=\frac{0±\sqrt{-24}}{2}$$

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.

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