Solve x⁴ + x³ - 28x² - 16x + 192 = 0
The solutions of x⁴ + x³ - 28x² - 16x + 192 = 0 are x = -4, -4, 3, 4 (repeated roots counted once each: -4, 3, 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 192, so the candidates are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, ±32, ±48, ±64, ±96, ±192.
- 2
Test the candidates until one gives 0. x = 3 is a root. Divide by (x − 3) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 3
x = 4 is a root. Divide by (x − 4) 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
\[ x^{4} + x^{3} - 28x^{2} - 16x + 192 = (x + 4)^{2}(x - 3)(x - 4) = 0 \]
\[ x^{4} + x^{3} - 28x^{2} - 16x + 192 = 0 \]
\[ \begin{array}{c|ccccc} 3 & 1 & 1 & -28 & -16 & 192 \\ & 0 & 3 & 12 & -48 & -192 \\ \hline & 1 & 4 & -16 & -64 & 0 \end{array} \]
\[ \begin{array}{c|cccc} 4 & 1 & 4 & -16 & -64 \\ & 0 & 4 & 32 & 64 \\ \hline & 1 & 8 & 16 & 0 \end{array} \]
\[ x^{2} + 8x + 16 = 0 \Rightarrow x = -4 \]
\[ x^{4} + x^{3} - 28x^{2} - 16x + 192 = (x + 4)^{2}(x - 3)(x - 4) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 150 | no |
| -1 | 180 | no |
| 2 | 72 | no |
| -2 | 120 | no |
| 3 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | -1 | -1 |
| Product of the roots | 192 | 192 |
Questions
What are the roots of x⁴ + x³ - 28x² - 16x + 192 = 0?
The solutions of x⁴ + x³ - 28x² - 16x + 192 = 0 are x = -4, -4, 3, 4 (repeated roots counted once each: -4, 3, 4).
How do you factor x⁴ + x³ - 28x² - 16x + 192?
(x + 4)(x − 3)(x − 4).
What is the sum of the roots?
-1 (it equals −1 by Vieta's formulas).