Solve x⁴ - 6x³ - 16x² + 150x - 225 = 0
The solutions of x⁴ - 6x³ - 16x² + 150x - 225 = 0 are x = -5, 3, 3, 5 (repeated roots counted once each: -5, 3, 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 -225, so the candidates are ±1, ±3, ±5, ±9, ±15, ±25, ±45, ±75, ±225.
- 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 = 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 - 25 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} - 6x^{3} - 16x^{2} + 150x - 225 = (x + 5)(x - 3)^{2}(x - 5) = 0 \]
\[ x^{4} - 6x^{3} - 16x^{2} + 150x - 225 = 0 \]
\[ \begin{array}{c|ccccc} 3 & 1 & -6 & -16 & 150 & -225 \\ & 0 & 3 & -9 & -75 & 225 \\ \hline & 1 & -3 & -25 & 75 & 0 \end{array} \]
\[ \begin{array}{c|cccc} 3 & 1 & -3 & -25 & 75 \\ & 0 & 3 & 0 & -75 \\ \hline & 1 & 0 & -25 & 0 \end{array} \]
\[ x^{2} - 25 = 0 \Rightarrow x = 5,\ -5 \]
\[ x^{4} - 6x^{3} - 16x^{2} + 150x - 225 = (x + 5)(x - 3)^{2}(x - 5) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | -96 | no |
| -1 | -384 | no |
| 3 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | 6 | 6 |
| Product of the roots | -225 | -225 |
Questions
What are the roots of x⁴ - 6x³ - 16x² + 150x - 225 = 0?
The solutions of x⁴ - 6x³ - 16x² + 150x - 225 = 0 are x = -5, 3, 3, 5 (repeated roots counted once each: -5, 3, 5).
How do you factor x⁴ - 6x³ - 16x² + 150x - 225?
(x + 5)(x − 3)(x − 5).
What is the sum of the roots?
6 (it equals −(-6) by Vieta's formulas).