Solve r^3+3r^2-20r-100=0

$$r ^ { 3 } + 3 r ^ { 2 } - 20 r - 100 = 0$$

Solve for r

$$r=5$$

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 -100 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±100,±50,±25,±20,±10,±5,±4,±2,±1$$

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.

$$r=5$$

By Factor theorem, r-k is a factor of the polynomial for each root k. Divide r^{3}+3r^{2}-20r-100 by r-5 to get r^{2}+8r+20. Solve the equation where the result equals to 0.

$$r^{2}+8r+20=0$$

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, 8 for b, and 20 for c in the quadratic formula.

$$r=\frac{-8±\sqrt{8^{2}-4\times 1\times 20}}{2}$$

Do the calculations.

$$r=\frac{-8±\sqrt{-16}}{2}$$

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

$$r\in \emptyset$$

List all found solutions.

$$r=5$$