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Mathematics 17 Online
OpenStudy (anonymous):

Explain, in complete sentences, how you would completely factor 20x2 - 28x - 48.

OpenStudy (saifoo.khan):

4 (x+1) (5 x-12)

OpenStudy (anonymous):

saifoo.khan, I know how to factor, but I just guess and check, is there some way to do it faster? Or not really?

OpenStudy (anonymous):

there's something called the ac factoring method

OpenStudy (saifoo.khan):

Epaema, sorry i didnt get wht u said! sorry!

OpenStudy (anonymous):

I'm so confuzzled, haha

OpenStudy (saifoo.khan):

Why?

OpenStudy (anonymous):

Never mind. Haha! :D

OpenStudy (anonymous):

I think I'm supposed to use the grouping method :/

OpenStudy (anonymous):

so it's called "the ac factoring method"?

OpenStudy (saifoo.khan):

thts also grouping!

OpenStudy (saifoo.khan):

thts kindda shortcut.

OpenStudy (anonymous):

Ok, so it's called "the ac factoring method"?

OpenStudy (saifoo.khan):

Yes.

OpenStudy (anonymous):

There are a lot of different methods. For this one I would look at the numbers first. It should be apparent that you can factor out a 4 from the numbers: \[20x^2-28x-48 \Rightarrow 4(5x^2-7x-12)\] The grouping comes into play because we notice here that 12 - 5 = 7, so we get: \[4(5x^2-7x-12) \Rightarrow 4(5x^2+5x-12x-12) \Rightarrow 4(5x(x+1)-12(x+1))\] \[4(x+1)(5x-12)\]

OpenStudy (anonymous):

er, that should be " ...we notice here that 5x - 12x = -7x, so we..." my bad.

OpenStudy (anonymous):

Thank you!

OpenStudy (anonymous):

I still don't get the grouping part..........

OpenStudy (saifoo.khan):

idk why didnt u.

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