Solve x³ - x² - 32x + 60 = 0
The solutions of x³ - x² - 32x + 60 = 0 are x = -6, 2, 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 60, so the candidates are ±1, ±2, ±3, ±4, ±5, ±6, ±10, ±12, ±15, ±20, ±30, ±60.
- 2
Test the candidates until one gives 0. x = 2 is a root. Divide by (x − 2) 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
\[ x^{3} - x^{2} - 32x + 60 = (x + 6)(x - 2)(x - 5) = 0 \]
\[ x^{3} - x^{2} - 32x + 60 = 0 \]
\[ \begin{array}{c|cccc} 2 & 1 & -1 & -32 & 60 \\ & 0 & 2 & 2 & -60 \\ \hline & 1 & 1 & -30 & 0 \end{array} \]
\[ x^{2} + x - 30 = 0 \Rightarrow x = 5,\ -6 \]
\[ x^{3} - x^{2} - 32x + 60 = (x + 6)(x - 2)(x - 5) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 28 | no |
| -1 | 90 | no |
| 2 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | 1 | 1 |
| Product of the roots | -60 | -60 |
Questions
What are the roots of x³ - x² - 32x + 60 = 0?
The solutions of x³ - x² - 32x + 60 = 0 are x = -6, 2, 5.
How do you factor x³ - x² - 32x + 60?
(x + 6)(x − 2)(x − 5).
What is the sum of the roots?
1 (it equals −(-1) by Vieta's formulas).