Solve x⁴ + 14x³ + 71x² + 154x + 120 = 0
The solutions of x⁴ + 14x³ + 71x² + 154x + 120 = 0 are x = -5, -4, -3, -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 120, so the candidates are ±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±40, ±60, ±120.
- 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 + 9x + 20 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} + 14x^{3} + 71x^{2} + 154x + 120 = (x + 5)(x + 4)(x + 3)(x + 2) = 0 \]
\[ x^{4} + 14x^{3} + 71x^{2} + 154x + 120 = 0 \]
\[ \begin{array}{c|ccccc} -2 & 1 & 14 & 71 & 154 & 120 \\ & 0 & -2 & -24 & -94 & -120 \\ \hline & 1 & 12 & 47 & 60 & 0 \end{array} \]
\[ \begin{array}{c|cccc} -3 & 1 & 12 & 47 & 60 \\ & 0 & -3 & -27 & -60 \\ \hline & 1 & 9 & 20 & 0 \end{array} \]
\[ x^{2} + 9x + 20 = 0 \Rightarrow x = -4,\ -5 \]
\[ x^{4} + 14x^{3} + 71x^{2} + 154x + 120 = (x + 5)(x + 4)(x + 3)(x + 2) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 360 | no |
| -1 | 24 | no |
| 2 | 840 | no |
| -2 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | -14 | -14 |
| Product of the roots | 120 | 120 |
Questions
What are the roots of x⁴ + 14x³ + 71x² + 154x + 120 = 0?
The solutions of x⁴ + 14x³ + 71x² + 154x + 120 = 0 are x = -5, -4, -3, -2.
How do you factor x⁴ + 14x³ + 71x² + 154x + 120?
(x + 5)(x + 4)(x + 3)(x + 2).
What is the sum of the roots?
-14 (it equals −14 by Vieta's formulas).