Posted by Jasmine20 on Wednesday, January 24, 2007 at 10:42pm.
Okay now i need explanation how to do these types of problems.
Directions: Divide
Problem:
(7x^5y^521x^4y^4+14x^3y^3)/(7x^3y^3)
Realize that you have 7x^3y^3 on the bottom. Now in order to get rid of that, you want to have the same thing in the top, multiplied by something else. That way, the 7x^3y^3 on the top and bottom will cross out, because any number over itself is 1. So, take out 7x^3y^3 from the numerator:
[7x^3y^3(x^2y^23xy+2)]/(7x^3y^3)
Now once you cross out the (7x^3y^3) from the fraction, you are left with x^2y^23xy+2. So, to factor this, you need the beginnings of the parentheses to be xy, so when you multiply xy times xy you will get x^2y^2.
(xy )(xy )
Now, you know that the ends of the parentheses must multiply to 2, so they will be 2 and 1, but we don't know if they are positive and negative yet.
(xy 2)(xy 1)
Now, you want the outer terms multiplied plus the inner terms multiplied to equal negative 3. So, if the 2 and 1 were positive, we'd end up with +3 and we want 3, so we know the signs inside must be negative:
(xy2)(xy1)
where did you get:
[7x^3y^3(x^2y^23xy+2)]/(7x^3y^3
the 3xy+2
I took out 7x^3y^3 from everything in the numerator. So, from the 7x^5y^5, I took out 7x^3y^3 and separated it from x^2y^2, because if you multiply those 2, you will end up with the original 7x^5y^5. I did the same for 21x^4y^4. If you remove from that 7x^3y^3 (basically, what will you multiply by 7x^3y^3 to get back to 21x^4y^4?), you will get 3xy. For the last part, 7x^3y^3 times 2 is the original, 14x^3y^3. Wow, that sounds way more confusing than it is. If you ask your teacher to show you it is much easier written and pointed out than me trying to explain it on here.

math,algebra,help  Tina, Monday, April 25, 2011 at 6:45pm
3y divided by 7
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