Factor completely.
\[18x^3y-33x^2y^2-30xy^3\]
is there something that we could factor out of this?
maybe something that all the terms have in common?
yes the xy and hmm i guess the number 3?
very good, if we factor those out we will get: 3xy(6x^2-11xy-10y^2)
now, we have to see if that quadratic onnthe inside can be factored, what do you think?
hmmm im not sure i thought 2 but that is wrong.
3xy(6x^2-11xy-10y^2) Well, we can use a bit of guess and check: lets say we start off with: (3x+?)(2x+?), since these two particular terms well give us 6x^2 okay
ok got it so far
Sorry, one of them has to have a minus sign,now (3x+?)(2x-?)=6x^2-3x?+2x?-?^2
now, the reason i did all of that is because we need to find numbers that will give us what we want, namely 6x^2-11xy-10y^2
So what we factor with has to have a minus sign?
yes, because if you notice the orginal quadratic has :6x^2-11xy-10y^2, it has two minus signs. So we need a minus sign in one of our terms. Now let me as you what comnbination of number well give us -10y^2
thats i to say, what numbers when ,multiplied together give us -10y^2
hmmm -2 and 5? right?
yes, in fact it s -2y and 5y, right, since we also want that y^2 part
oh yeah sorry forgot that part.
now we have (3x+?,)(2x-?), but we have found the numbers that should replace the quetion marks, can you postion those numbers so that we get: 6x^2-11xy-10y^2
so wait is it 3x-2 and 2x+5?
well, remeber we found the two numbers to be 2y and 5y
oh yeah so 2y and 5y sorry haha
so, now you are saying that it should be : (3x-2y) and (2x+5y), does, that when you multiply it our, give you 6x^2-11xy-10y^2
i think so cause 3*2=6 and -2*5=-10 and how do you get the 11xy
its from the two middle terms
oh yeah forgot bout foil
now, you have the correct terms. But just make sure you get the result you want.
the answer should look something like this ?xy(?x+?y)(?x-?y)
The answer, when completlely, factored looks like this: 3xy(2x-5y)(3x+3y)
hmmm i got it wrong but i have 2 more tries i wonder what is wrong.
I got it wrong.
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