Solve x³ - 6x² + 3x + 10 = 0
The solutions of x³ - 6x² + 3x + 10 = 0 are x = -1, 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 10, so the candidates are ±1, ±2, ±5, ±10.
- 2
Test the candidates until one gives 0. x = -1 is a root. Divide by (x + 1) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 3
What is left is the quadratic x² - 7x + 10 = 0. Solve it (factoring or the quadratic formula):
- 4
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{3} - 6x^{2} + 3x + 10 = (x + 1)(x - 2)(x - 5) = 0 \]
\[ x^{3} - 6x^{2} + 3x + 10 = 0 \]
\[ \begin{array}{c|cccc} -1 & 1 & -6 & 3 & 10 \\ & 0 & -1 & 7 & -10 \\ \hline & 1 & -7 & 10 & 0 \end{array} \]
\[ x^{2} - 7x + 10 = 0 \Rightarrow x = 5,\ 2 \]
\[ x^{3} - 6x^{2} + 3x + 10 = (x + 1)(x - 2)(x - 5) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 8 | no |
| -1 | 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 | -10 | -10 |
Questions
What are the roots of x³ - 6x² + 3x + 10 = 0?
The solutions of x³ - 6x² + 3x + 10 = 0 are x = -1, 2, 5.
How do you factor x³ - 6x² + 3x + 10?
(x + 1)(x − 2)(x − 5).
What is the sum of the roots?
6 (it equals −(-6) by Vieta's formulas).