View Full Version : factoring polynomials and trinomials.

skullcandy

February 22nd, 2009, 06:59 PM

I need someone to completely guide me through factoring polynomials and trinomials. My current teacher is an idiot and only got the job because of her boobs, and the substitute we had last week did a better job teaching than the actual math teacher.

Alright, guide me through everything on this problem.

12a^3-9a^2+20a-15

^ means to the exponent of

crazyassmetalhead

February 22nd, 2009, 07:10 PM

my teacher looks like little nicky and may possibly be retarded so i feel your pain.

though i don't need the credit in that class to graduate so i didn't bother trying to learn this shit, so i can't really help. though you can just google factoring polynomials and trinomials for a tutorial.

katt

February 22nd, 2009, 07:41 PM

what math level is this?

crazyassmetalhead

February 22nd, 2009, 07:48 PM

it's algebra

katt

February 22nd, 2009, 08:03 PM

what level though...there are multiple levels of algebra in ca

marcraft

February 22nd, 2009, 08:39 PM

grade 10 pre cal i think , i only go to school 2 days a week so im taking consumer which i can pass on 2 days a week.

skullcandy

February 22nd, 2009, 09:12 PM

I don't really need a calculator for this, just a simple explanation of where these numbers are getting pulled from, otherwise it just looks like they're coming out of your ass.

Sionyx

February 23rd, 2009, 03:26 AM

I'm not good at putting my shit into words, so here are the steps and what i did:

Grouped shit together, nothing has changed:

[12x^3 - 9x^2] + [20x - 15]

Removed common factors within the groups:

3x^2[4x - 3] + 5[4x - 3]

From this, we use the common factor of [4x - 3] and multiply it by coefficients of the factor 3x^2 and 5:

(3x^2 + 5)(4x - 3)

Obscurum14

February 23rd, 2009, 06:20 PM

Could use the factor theorem, it's not the easiest esp. if the equation is complicated, but it works.

Basically, put in values like a=1, a=2, a=-1, a=-2 etc.

If the equation equals zero, then that value is a factor.

freetibet

February 23rd, 2009, 08:44 PM

I'm not good at putting my shit into words, so here are the steps and what i did:

Grouped shit together, nothing has changed:

[12x^3 - 9x^2] + [20x - 15]

Removed common factors within the groups:

3x^2[4x - 3] + 5[4x - 3]

From this, we use the common factor of [4x - 3] and multiply it by coefficients of the factor 3x^2 and 5:

(3x^2 + 5)(4x - 3)

Wouldn't it be (3x^2 + 5)(4x - 3)^2 ???

Obscurum14

February 23rd, 2009, 09:02 PM

No it wouldn't, expand out (3x^2 + 5)(4x - 3) and you get the initial equation.

freetibet

February 23rd, 2009, 09:45 PM

oh ok I get it.

Cuz you factor out the (4x - 3)

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