Solve x⁴ - 12x³ + 37x² + 30x - 200 = 0
The solutions of x⁴ - 12x³ + 37x² + 30x - 200 = 0 are x = -2, 4, 5, 5 (repeated roots counted once each: -2, 4, 5).
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 -200, so the candidates are ±1, ±2, ±4, ±5, ±8, ±10, ±20, ±25, ±40, ±50, ±100, ±200.
- 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 = 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 - 10x + 25 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} - 12x^{3} + 37x^{2} + 30x - 200 = (x + 2)(x - 4)(x - 5)^{2} = 0 \]
\[ x^{4} - 12x^{3} + 37x^{2} + 30x - 200 = 0 \]
\[ \begin{array}{c|ccccc} -2 & 1 & -12 & 37 & 30 & -200 \\ & 0 & -2 & 28 & -130 & 200 \\ \hline & 1 & -14 & 65 & -100 & 0 \end{array} \]
\[ \begin{array}{c|cccc} 4 & 1 & -14 & 65 & -100 \\ & 0 & 4 & -40 & 100 \\ \hline & 1 & -10 & 25 & 0 \end{array} \]
\[ x^{2} - 10x + 25 = 0 \Rightarrow x = 5 \]
\[ x^{4} - 12x^{3} + 37x^{2} + 30x - 200 = (x + 2)(x - 4)(x - 5)^{2} \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | -144 | no |
| -1 | -180 | no |
| 2 | -72 | no |
| -2 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | 12 | 12 |
| Product of the roots | -200 | -200 |
Questions
What are the roots of x⁴ - 12x³ + 37x² + 30x - 200 = 0?
The solutions of x⁴ - 12x³ + 37x² + 30x - 200 = 0 are x = -2, 4, 5, 5 (repeated roots counted once each: -2, 4, 5).
How do you factor x⁴ - 12x³ + 37x² + 30x - 200?
(x + 2)(x − 4)(x − 5).
What is the sum of the roots?
12 (it equals −(-12) by Vieta's formulas).