Which is a factor of 6x3y + 6xy – 12x2– 12?
x + 2
xy – 1
x2 + 1
x2 – 2
How do I do this?
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rebeccaxhawaii (rebeccaxhawaii):
can you tell us which ones are exponents
OpenStudy (isaiah.feynman):
Hang on Rebecca. :P
rebeccaxhawaii (rebeccaxhawaii):
hang on isaiah
OpenStudy (adrianna.gongora):
\[6x ^{3} y + 6xy – 12x ^{2}– 12\]
x + 2
xy – 1
\[x ^{2}+1\]
\[x ^{2}-2\]
rebeccaxhawaii (rebeccaxhawaii):
so now we shall divide
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OpenStudy (adrianna.gongora):
Divide what ? c:
rebeccaxhawaii (rebeccaxhawaii):
hold on lemme write it out first
OpenStudy (adrianna.gongora):
okay :)
rebeccaxhawaii (rebeccaxhawaii):
so what can we easily factor out?
OpenStudy (adrianna.gongora):
6x
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OpenStudy (adrianna.gongora):
or 6xy?
rebeccaxhawaii (rebeccaxhawaii):
look at each term does all of them have x or xy??
OpenStudy (adrianna.gongora):
ohh okay so 6x
rebeccaxhawaii (rebeccaxhawaii):
does 12 have x in it??
OpenStudy (isaiah.feynman):
@phi @Kainui
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OpenStudy (adrianna.gongora):
Only one of them but it also has an exponent... oohhh okay so add 6x^3 and -12x^2
rebeccaxhawaii (rebeccaxhawaii):
just start off by dividing all the terms by 6. then we can factor the rest.
OpenStudy (adrianna.gongora):
so I have \[x ^{3}y+xy-2x ^{2}-2\]
OpenStudy (adrianna.gongora):
am I right? @rebeccaxhawaii
OpenStudy (adrianna.gongora):
@mathmale ?
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OpenStudy (mathmale):
have you actually chosen one of the four possible answers? Have you demonstrated that that answer (factor) actually divides into the given expression with no remainder?
OpenStudy (adrianna.gongora):
._.)/ what?
OpenStudy (adrianna.gongora):
I was going step by step ._.
OpenStudy (adrianna.gongora):
@Isaiah.Feynman
OpenStudy (anonymous):
this is probably one of those goofy "factor by grouping" questions
is it clear that there is a common factor of 6 in each term? factor that out first
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OpenStudy (anonymous):
\[6(x^3y+xy-2x^2-2)\]
OpenStudy (adrianna.gongora):
;-; so would it be wait that would put me back to where I started .-.
OpenStudy (anonymous):
the first two terms in inside the parentheses have a common factor of \(xy\)\[6(\color{red}{x^3y+xy}-2x^2-2)\]
OpenStudy (anonymous):
and the last two terms have a common factor of \(-2\)
OpenStudy (adrianna.gongora):
but theres a exponent
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OpenStudy (anonymous):
relax, it will all work out
OpenStudy (adrianna.gongora):
;-; ok
OpenStudy (anonymous):
\[6(2y(x^2+1)-2(x^2+1))\]
OpenStudy (anonymous):
now each of those terms inside the parentheses have a common factor of \(x^2+1\) so we can "factor it out" as the math teachers day, and end with \[6(x^2+1)(xy-2)\]
OpenStudy (anonymous):
and there it is displayed in factored form
imagine how they cooked up such a silly question !
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OpenStudy (adrianna.gongora):
x + 2
xy – 1
x2 + 1
x2 – 2
these are the options for the answers