Solve x⁴ + 18x³ + 120x² + 350x + 375 = 0
The solutions of x⁴ + 18x³ + 120x² + 350x + 375 = 0 are x = -5, -5, -5, -3 (repeated roots counted once each: -5, -3).
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 375, so the candidates are ±1, ±3, ±5, ±15, ±25, ±75, ±125, ±375.
- 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 = -5 is a root. Divide by (x + 5) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 4
What is left is the quadratic x² + 10x + 25 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} + 18x^{3} + 120x^{2} + 350x + 375 = (x + 5)^{3}(x + 3) = 0 \]
\[ x^{4} + 18x^{3} + 120x^{2} + 350x + 375 = 0 \]
\[ \begin{array}{c|ccccc} -3 & 1 & 18 & 120 & 350 & 375 \\ & 0 & -3 & -45 & -225 & -375 \\ \hline & 1 & 15 & 75 & 125 & 0 \end{array} \]
\[ \begin{array}{c|cccc} -5 & 1 & 15 & 75 & 125 \\ & 0 & -5 & -50 & -125 \\ \hline & 1 & 10 & 25 & 0 \end{array} \]
\[ x^{2} + 10x + 25 = 0 \Rightarrow x = -5 \]
\[ x^{4} + 18x^{3} + 120x^{2} + 350x + 375 = (x + 5)^{3}(x + 3) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 864 | no |
| -1 | 128 | no |
| 3 | 3072 | no |
| -3 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | -18 | -18 |
| Product of the roots | 375 | 375 |
Questions
What are the roots of x⁴ + 18x³ + 120x² + 350x + 375 = 0?
The solutions of x⁴ + 18x³ + 120x² + 350x + 375 = 0 are x = -5, -5, -5, -3 (repeated roots counted once each: -5, -3).
How do you factor x⁴ + 18x³ + 120x² + 350x + 375?
(x + 5)(x + 3).
What is the sum of the roots?
-18 (it equals −18 by Vieta's formulas).