# Thread: polynomial factoring

1. ## polynomial factoring

2. Originally Posted by algebra2
What are you asking?

The leading coefficient of the result of multiplying out your factorisation is 9, so it can't be right.

RonL

3. Hello, algebra2!

Your final factor is correct, but . . .

Your factoring is: . $3x^4 - 4x^3 + 4x^2 - 4x + 1 \;=\;(x-1)\left(x-\frac{1}{3}\right)3(x^2+1)$

The preferred form is: . $(x-1)(3x-1)(x^2+1)$

If complex numbers are allowed,
. . . that quadratic also factors: . $(x-1)(3x-1)(x-i)(x+i)$

4. 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 . $(x-1)\left(x-\frac{1}{3}\right)3(x^2+1)$

so when you got:

step 2. $(x-1)(3x-1)(x^2+1)$

what happened to the "3" in front of (x^2+1) in step 1?

5. Originally Posted by algebra2
what happened to the "3" in front of (x^2+1) in step 1?
It was multiplied into this term: $(x-\frac{1}{3})$

6. 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)?

7. Originally Posted by algebra2
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)?
There is no such rule. $x+\frac{a}{b} \ne bx+a$

This is what Soroban wrote for you: $3(x-\frac{1}{3}) = 3x - \frac{3}{3} \Rightarrow 3x - 1$

8. ok thanks.