Solve k^3+3k+4=0

$$k ^ { 3 } + 3 k + 4 = 0$$

Solve for k

$$k=-1$$

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 4 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±4,±2,±1$$

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}+3k+4 by k+1 to get k^{2}-k+4. Solve the equation where the result equals to 0.

$$k^{2}-k+4=0$$

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 4 for c in the quadratic formula.

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

Do the calculations.

$$k=\frac{1±\sqrt{-15}}{2}$$

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

$$k\in \emptyset$$

List all found solutions.

$$k=-1$$