Solve n^3-19n-30=0

$$n ^ { 3 } - 19 n - 30 = 0$$

Solve for n

$$n=-3$$

$$n=-2$$

$$n=5$$

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 -30 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±30,±15,±10,±6,±5,±3,±2,±1$$

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.

$$n=-2$$

By Factor theorem, n-k is a factor of the polynomial for each root k. Divide n^{3}-19n-30 by n+2 to get n^{2}-2n-15. Solve the equation where the result equals to 0.

$$n^{2}-2n-15=0$$

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 -15 for c in the quadratic formula.

$$n=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 1\left(-15\right)}}{2}$$

Do the calculations.

$$n=\frac{2±8}{2}$$

Solve the equation n^{2}-2n-15=0 when ± is plus and when ± is minus.

$$n=-3 n=5$$

List all found solutions.

$$n=-2 n=-3 n=5$$