# Thread: Factoring question using ac method

1. ## Factoring question using ac method

3x^2 -4x +1

The way I understand this is that you multiply 3 and 1 to get 3. I also understand that 3(1) or 1(3) would work by only if they added up to 4 and were able to be multiplied together to make 4 (bx). -3 -1 makes -4. But -3(-1) makes 3. With 3 being the constant I am really struggling to make this thing work.

The book gives the answer as (3x-1)(x-1)

Also: I have a similar problem with 2x^2 +11x -9: -2(9) and 2(-9) don't seem to be working either.

Could be I'm just fried because I'm tired but I'm stumped.

2. Originally Posted by Ingersoll
3x^2 -4x +1

The way I understand this is that you multiply 3 and 1 to get 3. I also understand that 3(1) or 1(3) would work by only if they added up to 4 and were able to be multiplied together to make 4 (bx). -3 -1 makes -4. But -3(-1) makes 3. With 3 being the constant I am really struggling to make this thing work.

The book gives the answer as (3x-1)(x-1)

Also: I have a similar problem with 2x^2 +11x -9: -2(9) and 2(-9) don't seem to be working either.

Could be I'm just fried because I'm tired but I'm stumped.
We assume there is a solution in integers.

Based on a = 3, we can reduce to one of these forms

(3x + u) (x + v)
(3x + u) (x - v)
(3x - u) (x + v)
(3x - u) (x - v)

for some positive integers u,v.

Based on c = 1, it's one of these forms

(3x + u) (x + v)
(3x - u) (x - v)

Based on negative b, it's this form

(3x - u) (x - v)

The only possibility for u and v are (1,1).

Next question: 2x^2 +11x -9

(2x + u) (x + v)
(2x + u) (x - v)
(2x - u) (x + v)
(2x - u) (x - v)

then

(2x + u) (x - v)
(2x - u) (x + v)

u,v = (1,9) or (3,3) or (9,1).

You can't get b=11 no matter what you try; this does not factor nicely in integers. (You can still factor using quadratic equation to find the two irrational roots.)