Solve 2x^5^+x^4-12x^3-12{x}^2+x+2=0

$$2 { x }^{ { 5 }^{ } } + { x }^{ 4 } -12 { x }^{ 3 } -12 { x }^{ 2 } +x+2 = 0$$

Solve for x

$$x=-2$$

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

$$x=\frac{3-\sqrt{5}}{2}\approx 0.381966011$$

$$x = \frac{\sqrt{5} + 3}{2} \approx 2.618033989$$

$$x=-1$$

Solution Steps

Calculate 5 to the power of 1 and get 5.

$$2x^{5}+x^{4}-12x^{3}-12x^{2}+x+2=0$$

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

$$±1,±2,±\frac{1}{2}$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{5}+x^{4}-12x^{3}-12x^{2}+x+2 by x+1 to get 2x^{4}-x^{3}-11x^{2}-x+2. Solve the equation where the result equals to 0.

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

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

$$±1,±2,±\frac{1}{2}$$

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.

$$x=-2$$

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

$$2x^{3}-5x^{2}-x+1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 1 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±\frac{1}{2},±1$$

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.

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

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

$$x^{2}-3x+1=0$$

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, -3 for b, and 1 for c in the quadratic formula.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$x=\frac{3±\sqrt{5}}{2}$$

Solve the equation x^{2}-3x+1=0 when ± is plus and when ± is minus.

$$x=\frac{3-\sqrt{5}}{2} x=\frac{\sqrt{5}+3}{2}$$

List all found solutions.

$$x=-1 x=-2 x=-\frac{1}{2} x=\frac{3-\sqrt{5}}{2} x=\frac{\sqrt{5}+3}{2}$$