Question

how the heck do I do this?!?!?!
(2xy-5x)^2

Answer 1

what are you trying to do?

Answer 2

Im trying to foil this expression, but I dont know what to do with the xy and the x

Answer 3

oooh okay so what you want to do is write that same equation out twice
so it will be
(2xy-5x)(2xy-5x)

Answer 4

and what we basically want to do is kind distribute

Answer 5

so 2xy times 2xy = 4x^2y^2
so now do that to the -5

Answer 6

so 2xy(-5) which is -10xy so so far we have 4x^2y^2-10xy

Answer 7

Well, can you do this? \((a-b)^2\)

If so, then it's just same process. You can let \(a = 2xy\) and \(b = 5x\).

Answer 8

now we want to do the -5 (2xy) which gives -10xy so we now have 4x^2y^2-10xy-10xy

Answer 9

how do I multiply xy by x?

Answer 10

what do you mean?

Answer 11

after foiling 2xy to the other 2xy, I have to multiply the 2xy by the -5x, and thats where I get stuck.

Answer 12

oh you just do it dont stop for nothing 2xy(-5x) you do -5 times 2 which gives you -10
how many x's are there? 2 so 10x^2 and now how many y's? 1 so it becomes 10x^2y

Answer 13

so when I foil xy and x, I have to combine the x's and leave the y?

Answer 14

If by "leaving" y, you mean do nothing, then yes.

You have \(xy\cdot x ~~~=~~~ x\cdot x\cdot y = x^2y\)

Answer 15

yes and the same would go if it was y y y x z it would be y^3xz....your basically counting the variables