Solve x⁴ + 2x³ - 16x² - 2x + 15 = 0
The solutions of x⁴ + 2x³ - 16x² - 2x + 15 = 0 are x = -5, -1, 1, 3.
How to work it out
- 1
A quartic has up to 4 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 15, so the candidates are ±1, ±3, ±5, ±15.
- 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
x = -1 is a root. Divide by (x + 1) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 4
What is left is the quadratic x² + 2x - 15 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} + 2x^{3} - 16x^{2} - 2x + 15 = (x + 5)(x + 1)(x - 1)(x - 3) = 0 \]
\[ x^{4} + 2x^{3} - 16x^{2} - 2x + 15 = 0 \]
\[ \begin{array}{c|ccccc} 1 & 1 & 2 & -16 & -2 & 15 \\ & 0 & 1 & 3 & -13 & -15 \\ \hline & 1 & 3 & -13 & -15 & 0 \end{array} \]
\[ \begin{array}{c|cccc} -1 & 1 & 3 & -13 & -15 \\ & 0 & -1 & -2 & 15 \\ \hline & 1 & 2 & -15 & 0 \end{array} \]
\[ x^{2} + 2x - 15 = 0 \Rightarrow x = 3,\ -5 \]
\[ x^{4} + 2x^{3} - 16x^{2} - 2x + 15 = (x + 5)(x + 1)(x - 1)(x - 3) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | -2 | -2 |
| Product of the roots | 15 | 15 |
Questions
What are the roots of x⁴ + 2x³ - 16x² - 2x + 15 = 0?
The solutions of x⁴ + 2x³ - 16x² - 2x + 15 = 0 are x = -5, -1, 1, 3.
How do you factor x⁴ + 2x³ - 16x² - 2x + 15?
(x + 5)(x + 1)(x − 1)(x − 3).
What is the sum of the roots?
-2 (it equals −2 by Vieta's formulas).