Solve x³ + 16x² + 85x + 150 = 0
The solutions of x³ + 16x² + 85x + 150 = 0 are x = -6, -5, -5 (repeated roots counted once each: -6, -5).
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 150, so the candidates are ±1, ±2, ±3, ±5, ±6, ±10, ±15, ±25, ±30, ±50, ±75, ±150.
- 2
Test the candidates until one gives 0. x = -5 is a root. Divide by (x + 5) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 3
What is left is the quadratic x² + 11x + 30 = 0. Solve it (factoring or the quadratic formula):
- 4
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{3} + 16x^{2} + 85x + 150 = (x + 6)(x + 5)^{2} = 0 \]
\[ x^{3} + 16x^{2} + 85x + 150 = 0 \]
\[ \begin{array}{c|cccc} -5 & 1 & 16 & 85 & 150 \\ & 0 & -5 & -55 & -150 \\ \hline & 1 & 11 & 30 & 0 \end{array} \]
\[ x^{2} + 11x + 30 = 0 \Rightarrow x = -5,\ -6 \]
\[ x^{3} + 16x^{2} + 85x + 150 = (x + 6)(x + 5)^{2} \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 252 | no |
| -1 | 80 | no |
| 2 | 392 | no |
| -2 | 36 | no |
| 3 | 576 | no |
| -3 | 12 | no |
| 5 | 1100 | no |
| -5 | 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 | -150 | -150 |
Questions
What are the roots of x³ + 16x² + 85x + 150 = 0?
The solutions of x³ + 16x² + 85x + 150 = 0 are x = -6, -5, -5 (repeated roots counted once each: -6, -5).
How do you factor x³ + 16x² + 85x + 150?
(x + 6)(x + 5).
What is the sum of the roots?
-16 (it equals −16 by Vieta's formulas).