Solve x³ - 17x² + 96x - 180 = 0
The solutions of x³ - 17x² + 96x - 180 = 0 are x = 5, 6, 6 (repeated roots counted once each: 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 -180, so the candidates are ±1, ±2, ±3, ±4, ±5, ±6, ±9, ±10, ±12, ±15, ±18, ±20, ±30, ±36, ±45, ±60, ±90, ±180.
- 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 - 12x + 36 = 0. Solve it (factoring or the quadratic formula):
- 4
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{3} - 17x^{2} + 96x - 180 = (x - 5)(x - 6)^{2} = 0 \]
\[ x^{3} - 17x^{2} + 96x - 180 = 0 \]
\[ \begin{array}{c|cccc} 5 & 1 & -17 & 96 & -180 \\ & 0 & 5 & -60 & 180 \\ \hline & 1 & -12 & 36 & 0 \end{array} \]
\[ x^{2} - 12x + 36 = 0 \Rightarrow x = 6 \]
\[ x^{3} - 17x^{2} + 96x - 180 = (x - 5)(x - 6)^{2} \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | -100 | no |
| -1 | -294 | no |
| 2 | -48 | no |
| -2 | -448 | no |
| 3 | -18 | no |
| -3 | -648 | no |
| 4 | -4 | no |
| -4 | -900 | no |
| 5 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | 17 | 17 |
| Product of the roots | 180 | 180 |
Questions
What are the roots of x³ - 17x² + 96x - 180 = 0?
The solutions of x³ - 17x² + 96x - 180 = 0 are x = 5, 6, 6 (repeated roots counted once each: 5, 6).
How do you factor x³ - 17x² + 96x - 180?
(x − 5)(x − 6).
What is the sum of the roots?
17 (it equals −(-17) by Vieta's formulas).