Solve x³ + x² - 16x + 20 = 0
The solutions of x³ + x² - 16x + 20 = 0 are x = -5, 2, 2 (repeated roots counted once each: -5, 2).
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 20, so the candidates are ±1, ±2, ±4, ±5, ±10, ±20.
- 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 + 3x - 10 = 0. Solve it (factoring or the quadratic formula):
- 4
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{3} + x^{2} - 16x + 20 = (x + 5)(x - 2)^{2} = 0 \]
\[ x^{3} + x^{2} - 16x + 20 = 0 \]
\[ \begin{array}{c|cccc} 2 & 1 & 1 & -16 & 20 \\ & 0 & 2 & 6 & -20 \\ \hline & 1 & 3 & -10 & 0 \end{array} \]
\[ x^{2} + 3x - 10 = 0 \Rightarrow x = 2,\ -5 \]
\[ x^{3} + x^{2} - 16x + 20 = (x + 5)(x - 2)^{2} \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 6 | no |
| -1 | 36 | 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 | -20 | -20 |
Questions
What are the roots of x³ + x² - 16x + 20 = 0?
The solutions of x³ + x² - 16x + 20 = 0 are x = -5, 2, 2 (repeated roots counted once each: -5, 2).
How do you factor x³ + x² - 16x + 20?
(x + 5)(x − 2).
What is the sum of the roots?
-1 (it equals −1 by Vieta's formulas).