http://img106.imageshack.us/img106/2200/polyds9.jpg
i used synthetic division.
http://img106.imageshack.us/img106/2200/polyds9.jpg
i used synthetic division.
Hello, algebra2!
Your final factor is correct, but . . .
Your factoring is: .$\displaystyle 3x^4 - 4x^3 + 4x^2 - 4x + 1 \;=\;(x-1)\left(x-\frac{1}{3}\right)3(x^2+1) $
The preferred form is: .$\displaystyle (x-1)(3x-1)(x^2+1)$
If complex numbers are allowed,
. . . that quadratic also factors: .$\displaystyle (x-1)(3x-1)(x-i)(x+i)$
i thought that for (x - 1/3) you put the denominator in front of the X and leave the numerator the way it is, so it becomes (3x - 1).
step 1 . $\displaystyle (x-1)\left(x-\frac{1}{3}\right)3(x^2+1)$
so when you got:
step 2. $\displaystyle (x-1)(3x-1)(x^2+1)$
what happened to the "3" in front of (x^2+1) in step 1?
what i'm trying to say is that isn't there a rule where:
(x - numerator/denominator)
can be rewritten as: (denominator times x - numerator)
for example:
(x - 1/4) ; (4x - 1)
(x - 2/5) ; (5x - 2)
(x - 1/3) ; (3x - 1)
and why was it multiplied to (x - 1/3) and not (x - 1)?