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12x^2-x-6 factors to (4x-3)(3x+2)
the problem is if you you use (4x+3)(3x-2)
if you distribute
you end up with 12x^2+x-6 not 12x^2-x-6
so it has to be (4x-3)(3x+2)
(a+b)(a-c) and (a-b)(a+c) do not distribute out to the same polynomials.
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)
when i redistribute it is wrong I see that, but even my homework shows the same process up until the final answer.
ok the problem is you are reversing the sign
12x+x-6 factors into (4x+3)(3x-2)
oops got that backwards
it factors into (4x-3)(3x+2)
(4x+3)(3x-2) would be the factors of 12x^2 + x -6
the difference is the x would be postive with these factors
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.
give me the full equation?
where is the +(-8x-6) coming from?
I got that from the factors for X (-8,9)
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)
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.
you have to have a - term i mean, lol
I see it now! Thank you