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Mathematics 7 Online
OpenStudy (adrianna.gongora):

Which is a factor of 6x3y + 6xy – 12x2– 12? x + 2 xy – 1 x2 + 1 x2 – 2 How do I do this?

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

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

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

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 ?

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

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

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 !

OpenStudy (adrianna.gongora):

x + 2 xy – 1 x2 + 1 x2 – 2 these are the options for the answers

OpenStudy (adrianna.gongora):

soo B?

OpenStudy (adrianna.gongora):

;-; omfg I'm just gonna go with the 2nd one.

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