Solve t^3-t+6=0

$$t ^ { 3 } - t + 6 = 0$$

Solve for t

$$t=-2$$

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

$$±6,±3,±2,±1$$

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.

$$t=-2$$

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

$$t^{2}-2t+3=0$$

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 1 for a, -2 for b, and 3 for c in the quadratic formula.

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

Do the calculations.

$$t=\frac{2±\sqrt{-8}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$t\in \emptyset$$

List all found solutions.

$$t=-2$$