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post the question please
what I would do is create a table like this |dw:1440551733478:dw|
the y+2 goes up top |dw:1440551751828:dw|
and the x+3 goes off to the side |dw:1440551769374:dw|
to fill this table out, we just multiply the headers example: 2 times 3 = 6 which goes in the second row, second column |dw:1440551792711:dw|
do you see how to fill out the rest?
kinda but the first one no
x times y is simply xy |dw:1440551935391:dw|
2 times x is equal to what?
and the last box?
now you just add up everything in the boxes \[\Large xy + 2x + 3y + 6\] there are no like terms to combine so we leave it as that
this all means \[\Large (x+3)(y+2) = xy + 2x + 3y + 6\]
3x 4y and 6
what do you mean?
thats all them added
but idk what to do with the 6
you cannot add 3x to 2y to get 5x or 5y or 5xy they aren't like terms
i added 3y and y
\(\Large xy + 2x + 3y + 6\) is as simplified as it gets
there is no y term other than the 3y the xy cannot be added to 3y to get 4y
so i dont need to use horizontal or vertical method?
you could if you want, but I prefer the table.
the horizontal or vertical method makes it hard to keep track of the individual terms whereas the table doesn't have that problem
so are there going to be other probs where they are more simplified?
yes, for example (x+2)(x-7) |dw:1440552711856:dw|
we now have like terms 2x and -7x they combine to -5x
so, (x+2)(x-7) = x^2 - 5x - 14