Solve x⁴ + 4x³ + 5x² + 2x = 0
The solutions of x⁴ + 4x³ + 5x² + 2x = 0 are x = -2, -1, -1, 0 (repeated roots counted once each: -2, -1, 0).
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 0 (it is 0, so x = 0 is a root and x factors out).
- 2
Test the candidates until one gives 0. x = 0 is a root. Divide by (x − 0) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 3
x = -1 is a root. Divide by (x + 1) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 4
What is left is the quadratic x² + 3x + 2 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} + 4x^{3} + 5x^{2} + 2x = (x + 2)(x + 1)^{2}x = 0 \]
\[ x^{4} + 4x^{3} + 5x^{2} + 2x = 0 \]
\[ \begin{array}{c|ccccc} 0 & 1 & 4 & 5 & 2 & 0 \\ & 0 & 0 & 0 & 0 & 0 \\ \hline & 1 & 4 & 5 & 2 & 0 \end{array} \]
\[ \begin{array}{c|cccc} -1 & 1 & 4 & 5 & 2 \\ & 0 & -1 & -3 & -2 \\ \hline & 1 & 3 & 2 & 0 \end{array} \]
\[ x^{2} + 3x + 2 = 0 \Rightarrow x = -1,\ -2 \]
\[ x^{4} + 4x^{3} + 5x^{2} + 2x = (x + 2)(x + 1)^{2}x \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 0 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | -4 | -4 |
| Product of the roots | 0 | 0 |
Questions
What are the roots of x⁴ + 4x³ + 5x² + 2x = 0?
The solutions of x⁴ + 4x³ + 5x² + 2x = 0 are x = -2, -1, -1, 0 (repeated roots counted once each: -2, -1, 0).
How do you factor x⁴ + 4x³ + 5x² + 2x?
(x + 2)(x + 1)x.
What is the sum of the roots?
-4 (it equals −4 by Vieta's formulas).