Solve 6x^3-7x^2-x+2=0

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

Solve for x

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

$$x=1$$

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

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

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

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 6x^{3}-7x^{2}-x+2 by x-1 to get 6x^{2}-x-2. Solve the equation where the result equals to 0.

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

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

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

Do the calculations.

$$x=\frac{1±7}{12}$$

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

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

List all found solutions.

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