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Mathematics 9 Online
OpenStudy (kidrah69):

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OpenStudy (kidrah69):

@zepdrix

zepdrix (zepdrix):

So what are we working on? :)

OpenStudy (kidrah69):

The same thing as yesterday ()_()

zepdrix (zepdrix):

mm i had another one ready.. lemme see if i can find it

zepdrix (zepdrix):

\(\Large\rm 5x^2-12x-9\) Did we do this one yet? Mmm I don't think so. Give it a try.

OpenStudy (kidrah69):

Ok first multiply so i get... -45 then the worst part ever >:( |dw:1431644924194:dw| can i pwease just find a list? c:

zepdrix (zepdrix):

why? you can't use a list on the test. that's a stupid question :(

OpenStudy (kidrah69):

I hate finding factors its too hard :( thats the part i screwed up T_T

zepdrix (zepdrix):

Maybe that's what we should practice then, not this whole problem, just the factors stuff. I'm going to list the factors of 60: \(\Large\rm 1,2,3,4,5,6,10,12,15,20,30,60\)

zepdrix (zepdrix):

What are the factors of 45, there are a lot less, it should be easier.

OpenStudy (kidrah69):

ok 1 x 45 = 45 9 x 5 = 45 thats all there is .-.

zepdrix (zepdrix):

no, that's not all of them.

zepdrix (zepdrix):

9 is a composite number

zepdrix (zepdrix):

so take a 3 out of the 9, give it to the 5

OpenStudy (kidrah69):

that would be 15 .-. 5 x 3 =15

zepdrix (zepdrix):

no.

OpenStudy (kidrah69):

1 3 5 9 15 45

zepdrix (zepdrix):

9 x 5 (3 x 3) x 5 3 x (3 x 5) 3 x 15

zepdrix (zepdrix):

It's not 3x5, u have another 3 after that

zepdrix (zepdrix):

so 3 and 15.... ok... -_-

OpenStudy (kidrah69):

lol wut? .-.

zepdrix (zepdrix):

\[\Large\rm \color{orangered}{9}\times5=45\]To find the other factors, you have to break up these composite numbers like this orange one.

OpenStudy (kidrah69):

Thats prime factorization right?

zepdrix (zepdrix):

\[\Large\rm \color{orangered}{(3\times3)}\times5=45\]And then group them differently,\[\Large\rm 3\times(3\times5)=45\]

zepdrix (zepdrix):

I dunno, it's something like that I guess.

zepdrix (zepdrix):

\[\Large\rm 3\times15=45\]

OpenStudy (kidrah69):

Ok oh :) so 1,3 5,9,15,45 Are the factors right?

zepdrix (zepdrix):

yes

zepdrix (zepdrix):

so finish it -_- nao .-.

OpenStudy (kidrah69):

|dw:1431645670997:dw|

zepdrix (zepdrix):

k c:

OpenStudy (kidrah69):

\[(5x^2-15x)(+3x-12)\] divide \[3x(x+3)~~~~~3(1x+3)\]

OpenStudy (kidrah69):

3x+3 x+3

OpenStudy (kidrah69):

DONE

zepdrix (zepdrix):

hmm not quite. let's see..

zepdrix (zepdrix):

it's really important that this plus be OUTSIDE the brackets,\[(5x^2-15x)+(3x-12)\]If I was your teacher, I would take some credit off for that :d

zepdrix (zepdrix):

3 from the second set? Looks good.\[(5x^2-15x)+3(x-4)\]

zepdrix (zepdrix):

You cannot take a 3 from the first step.

zepdrix (zepdrix):

Oh you made a boo boo earlier, why is there a -12 on the end. It's -9 D:

zepdrix (zepdrix):

\[\Large\rm 5x^2-12x-9\]

OpenStudy (kidrah69):

\[(5x^2-15x)~~(3x-9)\] first one needs to be divided by 5 \[5x(x-3)\] 2nd one by 3 \[3(x-3)\] then you get the answers... 5x+3 x-3

zepdrix (zepdrix):

Again, you forgot your plus sign, -1 points. Looks good besides that though! yay good job! \c:

OpenStudy (kidrah69):

TT_TT one point :c

zepdrix (zepdrix):

Let's try to find the factors of 72.

zepdrix (zepdrix):

No online list or table... just one pair at a time.

zepdrix (zepdrix):

what's 72 divided by 2?

OpenStudy (kidrah69):

36

zepdrix (zepdrix):

|dw:1431646510131:dw|

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