Solve x⁴ + 2x³ - 19x² - 32x + 48 = 0
The solutions of x⁴ + 2x³ - 19x² - 32x + 48 = 0 are x = -4, -3, 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 48, so the candidates are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, ±48.
- 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 = -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 - 16 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} + 2x^{3} - 19x^{2} - 32x + 48 = (x + 4)(x + 3)(x - 1)(x - 4) = 0 \]
\[ x^{4} + 2x^{3} - 19x^{2} - 32x + 48 = 0 \]
\[ \begin{array}{c|ccccc} 1 & 1 & 2 & -19 & -32 & 48 \\ & 0 & 1 & 3 & -16 & -48 \\ \hline & 1 & 3 & -16 & -48 & 0 \end{array} \]
\[ \begin{array}{c|cccc} -3 & 1 & 3 & -16 & -48 \\ & 0 & -3 & 0 & 48 \\ \hline & 1 & 0 & -16 & 0 \end{array} \]
\[ x^{2} - 16 = 0 \Rightarrow x = 4,\ -4 \]
\[ x^{4} + 2x^{3} - 19x^{2} - 32x + 48 = (x + 4)(x + 3)(x - 1)(x - 4) \]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 | 48 | 48 |
Questions
What are the roots of x⁴ + 2x³ - 19x² - 32x + 48 = 0?
The solutions of x⁴ + 2x³ - 19x² - 32x + 48 = 0 are x = -4, -3, 1, 4.
How do you factor x⁴ + 2x³ - 19x² - 32x + 48?
(x + 4)(x + 3)(x − 1)(x − 4).
What is the sum of the roots?
-2 (it equals −2 by Vieta's formulas).