1. ## [SOLVED] polynomial division

The answer for this one is m^2p + 4m^pn^2q - n^4q

2. Originally Posted by fishlord40

The answer for this one is m^2p + 4m^pn^2q - n^4q
note that $n^{4q} - 3m^pn^{2q} + 2m^{2p} = (n^{2q} - 2m^p)(n^{2q} - m^p)$, so that what you have is:

$\frac {5m^{3p}n^{2q} + 2m^{4p} - 13m^{2p}n^{4q} - n^{8q} + 7m^pn^{6q}}{(n^{2q} - 2m^p)(n^{2q} - m^p)}$

then divide $5m^{3p}n^{2q} + 2m^{4p} - 13m^{2p}n^{4q} - n^{8q} + 7m^pn^{6q}$ by $n^{2q} - m^p$ using polynomial long division.

then divide the new quotient by $n^{2q} - 2m^p$

3. Originally Posted by fishlord40

The answer for this one is m^2p + 4m^pn^2q - n^4q
$\frac{5m^{3p}n^{2q}+2m^{4p}-13m^{2p}n^{4q}-n^{8q}+7m^{p}n^{6q}}{n^{4q}-3m^{p}n^{2q}+2m^{2p}}$

use the long division but order the nominator like this

$-n^{8q}+7n^{6q}m^{p}-13n^{4q}m^{2p}+5n^{2q}m^{3p}+2m^{4p}$

try it

4. Originally Posted by fishlord40

The answer for this one is m^2p + 4m^pn^2q - n^4q
Start by arranging both polynomials in decreasing powers of m (or n if you prefer; but I've chosen m). The problem becomes: divide $2m^{4p} + 5m^{3p}n^{2q} - 13m^{2p}n^{4q} + 7m^pn^{6q} - n^{8q}$ by $2m^{2p} - 3m^pn^{2q} + n^{4q}$. Having done that, you will notice that m is always raised to a multiple of p, and n is always raised to a multiple of 2q. So it will save writing, and probably make the problem look easier, if we substitute x for $m^p$, and y for $n^{2q}$. Then the problem becomes: divide $2x^4 + 5x^3y - 13 x^2y^2 + 7xy^3 - y^4$ by $2x^2 - 3xy + y^2$.

Next, you might notice that $2x^2 - 3xy + y^2$ factorises as $2x^2 - 3xy + y^2 = (2x-y)(x-y)$. So the problem will be solved if we can divide $2x^4 + 5x^3y - 13 x^2y^2 + 7xy^3 - y^4$ by 2x–y, and then divide the result by x–y. You should be able to do that by whichever method you have been taught (probably either synthetic division or long division of polynomials). That will give you the answer, when you finally replace x by $m^p$ and y by $n^{2q}$.

5. Hello, fishlord40!

Divide: . $\left(5m^{3p}n^{2q} + 2m^{4p} - 13 ^{2p}n^{4q} - n^{8q} + 7m^pn^{6q}\right) \div \left(n^{4q} - 3m^pn^{2q} + 2m^{2p}\right)$

Answer: . $m^{2p} + 4m^pn^{2q} - n^{4q}$
How about Long Division, as Amer suggested ?

. . $\begin{array}{ccccccc}
& & & & m^{2p} & + 4m^pn^{2q} & - n^{4q} \\
& & --- & --- & --- & --- & --- \\
2m^{2p} - 3m^pq^{2q} + n^{4q} & ) & 2m^{4p} & +5m^{3p}n^{2q} & -13m^{2p}n^{4q} & +7m^pq^{6q} & - n^{8q} \\
& & 2m^{4p} & -3m^{3p}n^{2q} & +m^{2p}n^{4q} \\
& & --- & --- & --- \\
& & & 8m^{3p}n^{2q} & -14m^{2p}n^{4q} & +7m^pn^{6q} \\
\end{array}$

. . . . . . . . . . . . . . . . . . . . . . . $\begin{array}{cccc}8m^{3p}n^{2q} &- 12m^{2p}n^{4q} & + 4m^p q^{6q} \\
--- & --- & --- \\
& -2m^{2p}n^{4q} &+ 3m^pn^{6q} &- n^{8q} \\
& -2m^{2p}n^{4q} &+ 3m^pn^{6q} &- n^{8q} \\
& --- & --- & ---

\end{array}$

6. tnx for the help guys.

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