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Mathematics 14 Online
OpenStudy (anonymous):

Factoring polynomials by grouping?

OpenStudy (anonymous):

Can anyone explain this step by step? Please!! :)

OpenStudy (anonymous):

@phi @mathstudent55 anyone?

OpenStudy (anonymous):

Something like 7x^3 – 28x^2 + 3x – 12 that doesn't have a GCF

OpenStudy (zarkon):

\[ 7x^3 – 28x^2 + 3x – 12= (7x^3 – 28x^2) + (3x – 12)\]

OpenStudy (zarkon):

factor the two groups in parenthesis

OpenStudy (anonymous):

that makesss... 7x^2(x-4)+3(x-4) right?

OpenStudy (zarkon):

yes

OpenStudy (anonymous):

And where do we go from there?

OpenStudy (zarkon):

you have a common factor of x-4...factor that out

OpenStudy (anonymous):

wait like x-4(7x^2(x-4)+3(x-4)?

OpenStudy (zarkon):

no

OpenStudy (anonymous):

How?

OpenStudy (zarkon):

ac+bc=(a+b)c

OpenStudy (zarkon):

\[a=7x^2\] \[b=3\] \[c=x-4\]

OpenStudy (anonymous):

Oh so (7x^2+3)x-4?

OpenStudy (zarkon):

close the x-4 in parenthesis

OpenStudy (anonymous):

but with the communitive property of multiplication we can write it as (x-4)(7x^2+3) right?

OpenStudy (zarkon):

yes

OpenStudy (anonymous):

Oh I see, so we group the terms and force a common factor between the two!

OpenStudy (zarkon):

if possible ...yes

OpenStudy (anonymous):

and if not, its prime?

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