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March 29, 2015

March 29, 2015

Posted by **zach w** on Friday, November 30, 2012 at 9:30pm.

2) use synthetic division to factor X^2-2x^2-9x+18 completly

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**Reiny**, Friday, November 30, 2012 at 10:59pmif 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)

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