Factor the Polynomial: 12x^2-x-6
I got (4x+3)(3x-2) but my graded homework which indicates the answer is right shows (4x-3)(3x+2)

- anonymous

- katieb

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- anonymous

12x^2-x-6 factors to (4x-3)(3x+2)

- anonymous

the problem is if you you use (4x+3)(3x-2)

- anonymous

if you distribute

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## More answers

- anonymous

you end up with 12x^2+x-6
not 12x^2-x-6

- anonymous

so it has to be (4x-3)(3x+2)

- anonymous

(a+b)(a-c) and (a-b)(a+c) do not distribute out to the same polynomials.

- anonymous

I am not seeing where the signs change I will list my steps maybe you can point out where i am wrong
(12x^2-8x)+(9x-6)---> 4x(3x-2)+3(3x-2)--->(4x+3)(3x-2)

- anonymous

ok redistribute

- anonymous

when i redistribute it is wrong I see that, but even my homework shows the same process up until the final answer.

- anonymous

(4x+3)(3x-2)
12x+x-6

- anonymous

ok the problem is you are reversing the sign

- anonymous

12x+x-6 factors into
(4x+3)(3x-2)

- anonymous

not (4x-3)(3x+2)

- anonymous

oops got that backwards

- anonymous

it factors into (4x-3)(3x+2)

- anonymous

(4x+3)(3x-2)
would be the factors of 12x^2 + x -6

- anonymous

the difference is the x would be postive with these factors

- anonymous

If I played around with the equation I get:
(12x^2+9x)+(-8x-6)-->3x(4x+3)+2(-4x-3) which I am sure is wrong, somewhere I am screwing up with the signs.

- anonymous

give me the full equation?

- anonymous

12x^2-x-6

- anonymous

where is the +(-8x-6) coming from?

- anonymous

I got that from the factors for X (-8,9)

- anonymous

12x^2-x-6
(12x^2-8x)+(9x-6) <--- here is where you screwed up
12x^2+8x-9x-6
4x(3x+2)-3(3x+2)
(4x-3)(3x+2)

- anonymous

Reason being, is because it was originally "-x", so you can't have a + term in the middle while you are grouping the terms together. It has to be -, keeping it equal.

- anonymous

you have to have a - term i mean, lol

- anonymous

I see it now! Thank you

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