Solve 3x^3-14x^2+13x+6=0

$$3 x ^ { 3 } - 14 x ^ { 2 } + 13 x + 6 = 0$$

Solve for x

$$x=3$$

$$x=2$$

$$x=-\frac{1}{3}\approx -0.333333333$$

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 3. List all candidates \frac{p}{q}.

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

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 3x^{3}-14x^{2}+13x+6 by x-2 to get 3x^{2}-8x-3. Solve the equation where the result equals to 0.

$$3x^{2}-8x-3=0$$

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 3 for a, -8 for b, and -3 for c in the quadratic formula.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 3\left(-3\right)}}{2\times 3}$$

Do the calculations.

$$x=\frac{8±10}{6}$$

Solve the equation 3x^{2}-8x-3=0 when ± is plus and when ± is minus.

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

List all found solutions.

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