Solve k^3-7k+6=0

$$k ^ { 3 } - 7 k + 6 = 0$$

Solve for k

$$k=1$$

$$k=-3$$

$$k=2$$

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.

$$k=1$$

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

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

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

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

Do the calculations.

$$k=\frac{-1±5}{2}$$

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

$$k=-3 k=2$$

List all found solutions.

$$k=1 k=-3 k=2$$