First, we use the rational root theorem to determine any solutions of p(x). = x3 + 4x2 + x − 6
Factoring -6:
1
-1
2
-2
3
-3
6
-6
x = 1
p(1) = 1^3 + 4 * 1^2 + 1 - 6 = 6 - 6 = 0
x = 1 is a solution.
(x^3 + 4x^2 + x - 6) / (x - 1) =
x^3 / x = x^2
x^2 * (x - 1) = x^3 - x^2
x^3 + 4x^2 - x^3 + x^2 = 5x^2
5x^2 / x = 5x
5x * (x - 1) = 5x^2 - 5x
5x^2 + x - 5x^2 + 5x = 6x
6x / x = 6
6 * (x - 1) = 6x - 6
6x - 6 - 6x + 6 = 0
(x - 1) * (x^2 + 5x + 6)
x^2 + 5x + 6 factors to (x + 3) * (x + 2)
Factors:
(x - 1)
(x + 2)
(x + 3)
roots:
x = 1
x = -2
x = -3
