Solve x⁴ - 5x³ - 6x² + 32x + 32 = 0
The solutions of x⁴ - 5x³ - 6x² + 32x + 32 = 0 are x = -2, -1, 4, 4 (repeated roots counted once each: -2, -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 32, so the candidates are ±1, ±2, ±4, ±8, ±16, ±32.
- 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 = -2 is a root. Divide by (x + 2) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 4
What is left is the quadratic x² - 8x + 16 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} - 5x^{3} - 6x^{2} + 32x + 32 = (x + 2)(x + 1)(x - 4)^{2} = 0 \]
\[ x^{4} - 5x^{3} - 6x^{2} + 32x + 32 = 0 \]
\[ \begin{array}{c|ccccc} -1 & 1 & -5 & -6 & 32 & 32 \\ & 0 & -1 & 6 & 0 & -32 \\ \hline & 1 & -6 & 0 & 32 & 0 \end{array} \]
\[ \begin{array}{c|cccc} -2 & 1 & -6 & 0 & 32 \\ & 0 & -2 & 16 & -32 \\ \hline & 1 & -8 & 16 & 0 \end{array} \]
\[ x^{2} - 8x + 16 = 0 \Rightarrow x = 4 \]
\[ x^{4} - 5x^{3} - 6x^{2} + 32x + 32 = (x + 2)(x + 1)(x - 4)^{2} \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 54 | no |
| -1 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | 5 | 5 |
| Product of the roots | 32 | 32 |
Questions
What are the roots of x⁴ - 5x³ - 6x² + 32x + 32 = 0?
The solutions of x⁴ - 5x³ - 6x² + 32x + 32 = 0 are x = -2, -1, 4, 4 (repeated roots counted once each: -2, -1, 4).
How do you factor x⁴ - 5x³ - 6x² + 32x + 32?
(x + 2)(x + 1)(x − 4).
What is the sum of the roots?
5 (it equals −(-5) by Vieta's formulas).