Solve x⁴ + 13x³ + 63x² + 135x + 108 = 0
The solutions of x⁴ + 13x³ + 63x² + 135x + 108 = 0 are x = -4, -3, -3, -3 (repeated roots counted once each: -4, -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 108, so the candidates are ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±27, ±36, ±54, ±108.
- 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 + 7x + 12 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} + 13x^{3} + 63x^{2} + 135x + 108 = (x + 4)(x + 3)^{3} = 0 \]
\[ x^{4} + 13x^{3} + 63x^{2} + 135x + 108 = 0 \]
\[ \begin{array}{c|ccccc} -3 & 1 & 13 & 63 & 135 & 108 \\ & 0 & -3 & -30 & -99 & -108 \\ \hline & 1 & 10 & 33 & 36 & 0 \end{array} \]
\[ \begin{array}{c|cccc} -3 & 1 & 10 & 33 & 36 \\ & 0 & -3 & -21 & -36 \\ \hline & 1 & 7 & 12 & 0 \end{array} \]
\[ x^{2} + 7x + 12 = 0 \Rightarrow x = -3,\ -4 \]
\[ x^{4} + 13x^{3} + 63x^{2} + 135x + 108 = (x + 4)(x + 3)^{3} \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 320 | no |
| -1 | 24 | no |
| 2 | 750 | no |
| -2 | 2 | no |
| 3 | 1512 | no |
| -3 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | -13 | -13 |
| Product of the roots | 108 | 108 |
Questions
What are the roots of x⁴ + 13x³ + 63x² + 135x + 108 = 0?
The solutions of x⁴ + 13x³ + 63x² + 135x + 108 = 0 are x = -4, -3, -3, -3 (repeated roots counted once each: -4, -3).
How do you factor x⁴ + 13x³ + 63x² + 135x + 108?
(x + 4)(x + 3).
What is the sum of the roots?
-13 (it equals −13 by Vieta's formulas).