Solve x³ + 6x² - 25x - 150 = 0
The solutions of x³ + 6x² - 25x - 150 = 0 are x = -6, -5, 5.
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 -150, so the candidates are ±1, ±2, ±3, ±5, ±6, ±10, ±15, ±25, ±30, ±50, ±75, ±150.
- 2
Test the candidates until one gives 0. x = 5 is a root. Divide by (x − 5) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 3
What is left is the quadratic x² + 11x + 30 = 0. Solve it (factoring or the quadratic formula):
- 4
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{3} + 6x^{2} - 25x - 150 = (x + 6)(x + 5)(x - 5) = 0 \]
\[ x^{3} + 6x^{2} - 25x - 150 = 0 \]
\[ \begin{array}{c|cccc} 5 & 1 & 6 & -25 & -150 \\ & 0 & 5 & 55 & 150 \\ \hline & 1 & 11 & 30 & 0 \end{array} \]
\[ x^{2} + 11x + 30 = 0 \Rightarrow x = -5,\ -6 \]
\[ x^{3} + 6x^{2} - 25x - 150 = (x + 6)(x + 5)(x - 5) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | -168 | no |
| -1 | -120 | no |
| 2 | -168 | no |
| -2 | -84 | no |
| 3 | -144 | no |
| -3 | -48 | no |
| 5 | 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 | 150 | 150 |
Questions
What are the roots of x³ + 6x² - 25x - 150 = 0?
The solutions of x³ + 6x² - 25x - 150 = 0 are x = -6, -5, 5.
How do you factor x³ + 6x² - 25x - 150?
(x + 6)(x + 5)(x − 5).
What is the sum of the roots?
-6 (it equals −6 by Vieta's formulas).