Solve x⁴ - 3x³ - 18x² + 32x + 96 = 0
The solutions of x⁴ - 3x³ - 18x² + 32x + 96 = 0 are x = -3, -2, 4, 4 (repeated roots counted once each: -3, -2, 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 96, so the candidates are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, ±32, ±48, ±96.
- 2
Test the candidates until one gives 0. x = -2 is a root. Divide by (x + 2) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 3
x = -3 is a root. Divide by (x + 3) 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
\[ x^{4} - 3x^{3} - 18x^{2} + 32x + 96 = (x + 3)(x + 2)(x - 4)^{2} = 0 \]
\[ x^{4} - 3x^{3} - 18x^{2} + 32x + 96 = 0 \]
\[ \begin{array}{c|ccccc} -2 & 1 & -3 & -18 & 32 & 96 \\ & 0 & -2 & 10 & 16 & -96 \\ \hline & 1 & -5 & -8 & 48 & 0 \end{array} \]
\[ \begin{array}{c|cccc} -3 & 1 & -5 & -8 & 48 \\ & 0 & -3 & 24 & -48 \\ \hline & 1 & -8 & 16 & 0 \end{array} \]
\[ x^{2} - 8x + 16 = 0 \Rightarrow x = 4 \]
\[ x^{4} - 3x^{3} - 18x^{2} + 32x + 96 = (x + 3)(x + 2)(x - 4)^{2} \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 108 | no |
| -1 | 50 | no |
| 2 | 80 | no |
| -2 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | 3 | 3 |
| Product of the roots | 96 | 96 |
Questions
What are the roots of x⁴ - 3x³ - 18x² + 32x + 96 = 0?
The solutions of x⁴ - 3x³ - 18x² + 32x + 96 = 0 are x = -3, -2, 4, 4 (repeated roots counted once each: -3, -2, 4).
How do you factor x⁴ - 3x³ - 18x² + 32x + 96?
(x + 3)(x + 2)(x − 4).
What is the sum of the roots?
3 (it equals −(-3) by Vieta's formulas).