F.3 Factorization...爆急

consider a function

f(x)=x^3-7x+6

when x=1 , f(1)=1^3-7(1)+6=0

when x=2 , f(2)=2^3-7*2+6=0

By factor therorm , (x-1) and (x-2) are the factors of f(x)

so f(x)=(x-1)(x-2)(x+a) where a is a real number

f(0)=6 => 6=(0-1)(0-2)(0+a) =>a=3

so f(x)=(x-1)(x-2)(x+3)

Hence , x^3-7x+6=(x-1)(x-2)(x+3)//

use short division

see the image below to see how i did it

http://img229.imageshack.us/img229/6654/42162648.png

Let F(x) = x^3 -7x+6

first try to make f(x) =0

then u get f(1) =0

this means when f(x) is divided by x-1, it will be fully factorized

showing x-1 is a factor

put the coff on the right and put 1 on left when doing short division

then u get (x-1) (x^2+x-6)

factorize (x^2+x-6), then we have

f(x) = (x-1) (x-2) (x+3)

你肯定係 x^3 而唔係 x^2??
x^3 應該fact唔到喎