Solve x⁴ + 5x³ - 6x² - 32x + 32 = 0
The solutions of x⁴ + 5x³ - 6x² - 32x + 32 = 0 are x = -4, -4, 1, 2 (repeated roots counted once each: -4, 1, 2).
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
\[ x^{4} + 5x^{3} - 6x^{2} - 32x + 32 = (x + 4)^{2}(x - 1)(x - 2) = 0 \]
\[ x^{4} + 5x^{3} - 6x^{2} - 32x + 32 = 0 \]
\[ \begin{array}{c|ccccc} 1 & 1 & 5 & -6 & -32 & 32 \\ & 0 & 1 & 6 & 0 & -32 \\ \hline & 1 & 6 & 0 & -32 & 0 \end{array} \]
\[ \begin{array}{c|cccc} 2 & 1 & 6 & 0 & -32 \\ & 0 & 2 & 16 & 32 \\ \hline & 1 & 8 & 16 & 0 \end{array} \]
\[ x^{2} + 8x + 16 = 0 \Rightarrow x = -4 \]
\[ x^{4} + 5x^{3} - 6x^{2} - 32x + 32 = (x + 4)^{2}(x - 1)(x - 2) \]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 | -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 = -4, -4, 1, 2 (repeated roots counted once each: -4, 1, 2).
How do you factor x⁴ + 5x³ - 6x² - 32x + 32?
(x + 4)(x − 1)(x − 2).
What is the sum of the roots?
-5 (it equals −5 by Vieta's formulas).