# Math Help - divide

1. ## divide

divide the first expression by the second.
(12x^3-11x^y-12xy-25y^3)/(3x-5y)

i got lost when divisor has two variables.... thanks

yeah there is an error here.
(12x^3-11x^2y-12xy-25y^3)/(3x-5y)

and it is really 12xy there.
thanks

2. well there are a few ways to do this
first is synthetic division which they tought you in school but I don't quite remember
the next is just simply doing it one step at a time, it comes out to be the synthetic method I believe but this way you can see what you are doing

(12x^3-11x^y-12xy-25y^3)/(3x-5y)
12x^3, it is 4x^2 times the first term in the bottom
so replace 12x^3 with 4x^2(3x - 5y) + 20yx^2 , is it clear why I did that?
anyways now you have
(4x^2) + (20yx^2-11x^y-12xy-25y^3)/(3x-5y)
and you can keep going

hopefully this helps, I just woke up so I'm still not entirely lucid

3. Hello, princess_21!

I will assume there are typos in your problem . . .

$(12x^3-11x^{{\color{red}2}}y-12xy^{{\color{red}2}}-25y^3) \div (3x-5y)$

i got lost when divisor has two variables . . . really?

. . $\begin{array}{ccccccc}
& & & 4x^2 & +3xy & +y^2 \\
& & --- & --- & --- & --- \\
3x-5y & ) & 12x^3 & -11x^2y & -12xy^2 & -25y^3 \\
& & 12x^3 & -20x^2y \\
& & --- & --- \\
& & & 9x^2y & -12xy^2 \\
& & & 9x^2y & -15xy^2 \\
& & & -- & --- \\
& & & & 3xy^2 & -25y^3 \\
& & & & 3xy^2 & -5y^3 \\
& & & & --- & --- \end{array}$

. . . . . . . . . . . . . . . . . . . . . . . . . . $-20y^3$

Answer: . $4x^2 + 3xy + y^2 - \frac{20y^3}{3x-5y}$

4. Originally Posted by princess_21
divide the first expression by the second.
(12x^3-11x^y-12xy-25y^3)/(3x-5y)

i got lost when divisor has two variables.... thanks
$\frac{12x^3-11x^2y-12xy^2-25y^3}{3x-5y} =y^2\left(\frac{12a^3-11a^2-12a-25}{3a-5}\right)$ ,where $a = \frac{x}{y}$ Now you need to factorise in one variable... Hope you wont get lost

5. yeah there is an error here.
(12x^3-11x^2y-12xy-25y^3)/(3x-5y)

and it is really 12xy there.
thanks