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
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
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)
so start with the first term
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
Hello, princess_21!
I will assume there are typos in your problem . . .
$\displaystyle (12x^3-11x^{{\color{red}2}}y-12xy^{{\color{red}2}}-25y^3) \div (3x-5y)$
i got lost when divisor has two variables . . . really?
. . $\displaystyle \begin{array}{ccccccc}
& & & 4x^2 & +3xy & +y^2 \\
& & --- & --- & --- & --- \\
3x-5y & ) & 12x^3 & -11x^2y & -12xy^2 & -25y^3 \\
& & 12x^3 & -20x^2y \\
& & --- & --- \\
& & & 9x^2y & -12xy^2 \\
& & & 9x^2y & -15xy^2 \\
& & & -- & --- \\
& & & & 3xy^2 & -25y^3 \\
& & & & 3xy^2 & -5y^3 \\
& & & & --- & --- \end{array}$
. . . . . . . . . . . . . . . . . . . . . . . . . .$\displaystyle -20y^3$
Answer: .$\displaystyle 4x^2 + 3xy + y^2 - \frac{20y^3}{3x-5y} $