# Thread: [SOLVED] polynomial division

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 $\displaystyle n^{4q} - 3m^pn^{2q} + 2m^{2p} = (n^{2q} - 2m^p)(n^{2q} - m^p)$, so that what you have is:

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

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

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

3. Originally Posted by fishlord40

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

use the long division but order the nominator like this

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

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 $\displaystyle 2m^{4p} + 5m^{3p}n^{2q} - 13m^{2p}n^{4q} + 7m^pn^{6q} - n^{8q}$ by $\displaystyle 2m^{2p} - 3m^pn^{2q} + n^{4q}$. 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 $\displaystyle m^p$, and y for $\displaystyle n^{2q}$. Then the problem becomes: divide $\displaystyle 2x^4 + 5x^3y - 13 x^2y^2 + 7xy^3 - y^4$ by $\displaystyle 2x^2 - 3xy + y^2$.

Next, you might notice that $\displaystyle 2x^2 - 3xy + y^2$ factorises as $\displaystyle 2x^2 - 3xy + y^2 = (2x-y)(x-y)$. So the problem will be solved if we can divide $\displaystyle 2x^4 + 5x^3y - 13 x^2y^2 + 7xy^3 - y^4$ 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 $\displaystyle m^p$ and y by $\displaystyle n^{2q}$.

5. Hello, fishlord40!

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

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

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

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

6. tnx for the help guys.

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