Posted by zach w on Friday, November 30, 2012 at 9:30pm.
if f(c) = 0 , then x-c is a factor
for f(x) = x^2 - 2x^2 - 9x + 18
why do you have two x^2 terms, I will assume the first is x^3
If so, we don't need the factor theorem for this one, grouping is obvious
x^3 - 2x^2 - 9x + 18
= x^2(x-2) - 9(x-2)
= (x-2)(x^2 - 9)
= (x-2)(x+3)(x-3)
if you had not seen this, try x = ±1, ±2 , ±3 , that is factors of 18
f(1) ≠ 0
f(-1) ≠ 0
f(2) = 0 , yeahhhh, so x-2 is a factor
..
f(3) = 0 , so x-3 is a factor
f(-3) = 0 , so x+3 is a factor
since we have a cubic, there can only be a maximum of 3 algebraic factors
so as above
(x-2)(x+3)(x-3)