# 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?

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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:

a=1

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.