Solve x³ - 6x² - 25x + 150 = 0
The solutions of x³ - 6x² - 25x + 150 = 0 are x = -5, 5, 6.
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 - x - 30 = 0. Solve it (factoring or the quadratic formula):
- 4
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{3} - 6x^{2} - 25x + 150 = (x + 5)(x - 5)(x - 6) = 0 \]
\[ x^{3} - 6x^{2} - 25x + 150 = 0 \]
\[ \begin{array}{c|cccc} 5 & 1 & -6 & -25 & 150 \\ & 0 & 5 & -5 & -150 \\ \hline & 1 & -1 & -30 & 0 \end{array} \]
\[ x^{2} - x - 30 = 0 \Rightarrow x = 6,\ -5 \]
\[ x^{3} - 6x^{2} - 25x + 150 = (x + 5)(x - 5)(x - 6) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 120 | no |
| -1 | 168 | no |
| 2 | 84 | no |
| -2 | 168 | no |
| 3 | 48 | no |
| -3 | 144 | 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 = -5, 5, 6.
How do you factor x³ - 6x² - 25x + 150?
(x + 5)(x − 5)(x − 6).
What is the sum of the roots?
6 (it equals −(-6) by Vieta's formulas).