Solve x³ + 16x² + 84x + 144 = 0
The solutions of x³ + 16x² + 84x + 144 = 0 are x = -6, -6, -4 (repeated roots counted once each: -6, -4).
How to work it out
- 1
A cubic has up to 3 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 144, so the candidates are ±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±16, ±18, ±24, ±36, ±48, ±72, ±144.
- 2
Test the candidates until one gives 0. x = -4 is a root. Divide by (x + 4) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 3
What is left is the quadratic x² + 12x + 36 = 0. Solve it (factoring or the quadratic formula):
- 4
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{3} + 16x^{2} + 84x + 144 = (x + 6)^{2}(x + 4) = 0 \]
\[ x^{3} + 16x^{2} + 84x + 144 = 0 \]
\[ \begin{array}{c|cccc} -4 & 1 & 16 & 84 & 144 \\ & 0 & -4 & -48 & -144 \\ \hline & 1 & 12 & 36 & 0 \end{array} \]
\[ x^{2} + 12x + 36 = 0 \Rightarrow x = -6 \]
\[ x^{3} + 16x^{2} + 84x + 144 = (x + 6)^{2}(x + 4) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 245 | no |
| -1 | 75 | no |
| 2 | 384 | no |
| -2 | 32 | no |
| 3 | 567 | no |
| -3 | 9 | no |
| 4 | 800 | no |
| -4 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | -16 | -16 |
| Product of the roots | -144 | -144 |
Questions
What are the roots of x³ + 16x² + 84x + 144 = 0?
The solutions of x³ + 16x² + 84x + 144 = 0 are x = -6, -6, -4 (repeated roots counted once each: -6, -4).
How do you factor x³ + 16x² + 84x + 144?
(x + 6)(x + 4).
What is the sum of the roots?
-16 (it equals −16 by Vieta's formulas).