Solve x³ + 6x² - 15x - 100 = 0
The solutions of x³ + 6x² - 15x - 100 = 0 are x = -5, -5, 4 (repeated roots counted once each: -5, 4).
How to work it out
- 1
A cubic has up to 3 real roots. The rational root theorem says that any integer root of a polynomial with integer coefficients and leading coefficient 1 divides the constant term -100, so the candidates are ±1, ±2, ±4, ±5, ±10, ±20, ±25, ±50, ±100.
- 2
Test the candidates until one gives 0. x = 4 is a root. Divide by (x − 4) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 3
What is left is the quadratic x² + 10x + 25 = 0. Solve it (factoring or the quadratic formula):
- 4
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{3} + 6x^{2} - 15x - 100 = (x + 5)^{2}(x - 4) = 0 \]
\[ x^{3} + 6x^{2} - 15x - 100 = 0 \]
\[ \begin{array}{c|cccc} 4 & 1 & 6 & -15 & -100 \\ & 0 & 4 & 40 & 100 \\ \hline & 1 & 10 & 25 & 0 \end{array} \]
\[ x^{2} + 10x + 25 = 0 \Rightarrow x = -5 \]
\[ x^{3} + 6x^{2} - 15x - 100 = (x + 5)^{2}(x - 4) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | -108 | no |
| -1 | -80 | no |
| 2 | -98 | no |
| -2 | -54 | no |
| 4 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | -6 | -6 |
| Product of the roots | 100 | 100 |
Questions
What are the roots of x³ + 6x² - 15x - 100 = 0?
The solutions of x³ + 6x² - 15x - 100 = 0 are x = -5, -5, 4 (repeated roots counted once each: -5, 4).
How do you factor x³ + 6x² - 15x - 100?
(x + 5)(x − 4).
What is the sum of the roots?
-6 (it equals −6 by Vieta's formulas).