Solve 2x^4+x^3-8x^2-x+6=0

$$2 x ^ { 4 } + x ^ { 3 } - 8 x ^ { 2 } - x + 6 = 0$$

Solve for x

$$x=1$$

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$x=-2$$

$$x=-1$$

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

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

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{4}+x^{3}-8x^{2}-x+6 by x-1 to get 2x^{3}+3x^{2}-5x-6. Solve the equation where the result equals to 0.

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -6 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

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

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.

$$x=-1$$

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

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

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

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

Do the calculations.

$$x=\frac{-1±7}{4}$$

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

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

List all found solutions.

$$x=1 x=-1 x=-2 x=\frac{3}{2}$$