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