Solve x⁴ - 2x³ - 15x² + 32x - 16 = 0
The solutions of x⁴ - 2x³ - 15x² + 32x - 16 = 0 are x = -4, 1, 1, 4 (repeated roots counted once each: -4, 1, 4).
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 -16, so the candidates are ±1, ±2, ±4, ±8, ±16.
- 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 - 16 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} - 2x^{3} - 15x^{2} + 32x - 16 = (x + 4)(x - 1)^{2}(x - 4) = 0 \]
\[ x^{4} - 2x^{3} - 15x^{2} + 32x - 16 = 0 \]
\[ \begin{array}{c|ccccc} 1 & 1 & -2 & -15 & 32 & -16 \\ & 0 & 1 & -1 & -16 & 16 \\ \hline & 1 & -1 & -16 & 16 & 0 \end{array} \]
\[ \begin{array}{c|cccc} 1 & 1 & -1 & -16 & 16 \\ & 0 & 1 & 0 & -16 \\ \hline & 1 & 0 & -16 & 0 \end{array} \]
\[ x^{2} - 16 = 0 \Rightarrow x = 4,\ -4 \]
\[ x^{4} - 2x^{3} - 15x^{2} + 32x - 16 = (x + 4)(x - 1)^{2}(x - 4) \]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 | -16 | -16 |
Questions
What are the roots of x⁴ - 2x³ - 15x² + 32x - 16 = 0?
The solutions of x⁴ - 2x³ - 15x² + 32x - 16 = 0 are x = -4, 1, 1, 4 (repeated roots counted once each: -4, 1, 4).
How do you factor x⁴ - 2x³ - 15x² + 32x - 16?
(x + 4)(x − 1)(x − 4).
What is the sum of the roots?
2 (it equals −(-2) by Vieta's formulas).