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±\frac{1}{3},±\frac{2}{3},±1,±2,±\frac{1}{6},±\frac{1}{2}
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}.
x=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.
6x^{3}-5x^{2}-3x+2=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 6x^{4}-11x^{3}+2x^{2}+5x-2 by x-1 to get 6x^{3}-5x^{2}-3x+2. Solve the equation where the result equals to 0.
±\frac{1}{3},±\frac{2}{3},±1,±2,±\frac{1}{6},±\frac{1}{2}
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}.
x=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.
6x^{2}+x-2=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 6x^{3}-5x^{2}-3x+2 by x-1 to get 6x^{2}+x-2. Solve the equation where the result equals to 0.
x=\frac{-1±\sqrt{1^{2}-4\times 6\left(-2\right)}}{2\times 6}
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{-1±7}{12}
Do the calculations.
x=-\frac{2}{3} x=\frac{1}{2}
Solve the equation 6x^{2}+x-2=0 when ± is plus and when ± is minus.
x=1 x=-\frac{2}{3} x=\frac{1}{2}
List all found solutions.