## anonymous one year ago how the heck do I do this?!?!?! (2xy-5x)^2

1. anonymous

what are you trying to do?

2. anonymous

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

3. anonymous

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

4. anonymous

and what we basically want to do is kind distribute

5. anonymous

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

6. anonymous

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

7. geerky42

Well, can you do this? $$(a-b)^2$$ If so, then it's just same process. You can let $$a = 2xy$$ and $$b = 5x$$.

8. anonymous

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

9. anonymous

how do I multiply xy by x?

10. anonymous

what do you mean?

11. anonymous

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

12. anonymous

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

13. anonymous

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

14. geerky42

If by "leaving" y, you mean do nothing, then yes. You have $$xy\cdot x ~~~=~~~ x\cdot x\cdot y = x^2y$$

15. anonymous

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