apply factor theoremhow to apply factor theorem to find the three roots of P(x) = x³ − 2x² − 9x + 18, given that one root is 3?
Since 3 is a root of P(x), then according to the factor theorem, x − 3 is a factor. Therefore, we'll divide P(x) by x − 3 and we'll determine the other quadratic factor.
P(x) = (x-3)(ax^2+bx+c)
x^3 - 2x^2 - 9x + 18 = ax^3 + bx^2 + cx - 3ax^2 - 3bx - 3c
x^3 - 2x^2 - 9x + 18 = ax^3 + x^2*(b-3a) + x*(c - 3b) - 3c
Comparing, we'll get:
b - 3a = -2
b - 3 = -2
b = 1
c - 3b = -9
c - 3 = -9
c = -6
-3c = 18
c = -6
The quadratic factor is: x^2 + x -6
The roots of x^2 + x -6 are x = -3 and x = 2
The factor theorem applied to the given polynomial is:
P(x) = (x-2)(x+3)(x-3)
The reminder is zero because x=3 is the root of P(x), so x - 3 is a factor.