Solve (2x^4+3x^3-11x^2+2x+4)=0

$$(2 { x }^{ 4 } +3 { x }^{ 3 } -11 { x }^{ 2 } +2x+4)=0$$

Solve for x (complex solution)

$$x=-\frac{1}{2}=-0.5$$

$$x=1$$

$$x=-\left(\sqrt{5}+1\right)\approx -3.236067977$$

$$x=\sqrt{5}-1\approx 1.236067977$$

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

$$±2,±4,±1,±\frac{1}{2}$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{4}+3x^{3}-11x^{2}+2x+4 by x-1 to get 2x^{3}+5x^{2}-6x-4. Solve the equation where the result equals to 0.

$$2x^{3}+5x^{2}-6x-4=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -4 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±2,±4,±1,±\frac{1}{2}$$

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.

$$x=-\frac{1}{2}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}+5x^{2}-6x-4 by 2\left(x+\frac{1}{2}\right)=2x+1 to get x^{2}+2x-4. Solve the equation where the result equals to 0.

$$x^{2}+2x-4=0$$

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, 2 for b, and -4 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{-2±2\sqrt{5}}{2}$$

Solve the equation x^{2}+2x-4=0 when ± is plus and when ± is minus.

$$x=-\sqrt{5}-1 x=\sqrt{5}-1$$

List all found solutions.

$$x=1 x=-\frac{1}{2} x=-\sqrt{5}-1 x=\sqrt{5}-1$$

Solve for x

$$x=\sqrt{5}-1\approx 1.236067977$$

$$x=-\sqrt{5}-1\approx -3.236067977$$

$$x=1$$

$$x=-\frac{1}{2}=-0.5$$

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

$$±2,±4,±1,±\frac{1}{2}$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{4}+3x^{3}-11x^{2}+2x+4 by x-1 to get 2x^{3}+5x^{2}-6x-4. Solve the equation where the result equals to 0.

$$2x^{3}+5x^{2}-6x-4=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -4 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±2,±4,±1,±\frac{1}{2}$$

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.

$$x=-\frac{1}{2}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}+5x^{2}-6x-4 by 2\left(x+\frac{1}{2}\right)=2x+1 to get x^{2}+2x-4. Solve the equation where the result equals to 0.

$$x^{2}+2x-4=0$$

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, 2 for b, and -4 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{-2±2\sqrt{5}}{2}$$

Solve the equation x^{2}+2x-4=0 when ± is plus and when ± is minus.

$$x=-\sqrt{5}-1 x=\sqrt{5}-1$$

List all found solutions.

$$x=1 x=-\frac{1}{2} x=-\sqrt{5}-1 x=\sqrt{5}-1$$