Solve x³ + 4x² - 35x - 150 = 0
The solutions of x³ + 4x² - 35x - 150 = 0 are x = -5, -5, 6 (repeated roots counted once each: -5, 6).
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 - x - 30 = 0. Solve it (factoring or the quadratic formula):
- 4
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{3} + 4x^{2} - 35x - 150 = (x + 5)^{2}(x - 6) = 0 \]
\[ x^{3} + 4x^{2} - 35x - 150 = 0 \]
\[ \begin{array}{c|cccc} -5 & 1 & 4 & -35 & -150 \\ & 0 & -5 & 5 & 150 \\ \hline & 1 & -1 & -30 & 0 \end{array} \]
\[ x^{2} - x - 30 = 0 \Rightarrow x = 6,\ -5 \]
\[ x^{3} + 4x^{2} - 35x - 150 = (x + 5)^{2}(x - 6) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | -180 | no |
| -1 | -112 | no |
| 2 | -196 | no |
| -2 | -72 | no |
| 3 | -192 | no |
| -3 | -36 | no |
| 5 | -100 | no |
| -5 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | -4 | -4 |
| Product of the roots | 150 | 150 |
Questions
What are the roots of x³ + 4x² - 35x - 150 = 0?
The solutions of x³ + 4x² - 35x - 150 = 0 are x = -5, -5, 6 (repeated roots counted once each: -5, 6).
How do you factor x³ + 4x² - 35x - 150?
(x + 5)(x − 6).
What is the sum of the roots?
-4 (it equals −4 by Vieta's formulas).