# Thread: Polynomial Division with Variables n and m

1. ## Polynomial Division with Variables n and m

When 10x^3+mx^2-x+10 is divided by 5x-3, the quotient is 2x^2+nx-2 and the remainder is 4. Find the values of m and n.

The quotient multiplied by the divisor added to the remainder should equal the dividend, so:

$10x^3+mx^2-x+10=(5x-3)(2x^2+nx-2)+4$

Expand the right side:

$10x^3+mx^2-x+10=5nx^2-3nx+10x^3-6x^2-10x+10$

Cancel Terms:

$mx^2-x=5nx^2-3nx-6x^2-10x$

Factor:

$x(mx-1)=x(5nx-3n-6x-10)$

Cancel:

$mx-1=5nx-3n-6x-10$

Isolate m:

$m=\frac{5nx-3n-6x-9}{x}$

Sub m into initial equation to solve for n:

$10x^3+(\frac{5nx-3n-6x-9}{x})x^2-x+10=5nx^2-3nx+10x^3-6x^2-10x+10$

Expand:

$10x^3+5nx^2-3nx-6x^2-9x-x+10=5nx^2-3nx+10x^3-6x^2-10x+10$

Collect Like Terms:

$10x^3+5nx^2-3nx-6x^2-10x+10=5nx^2-3nx+10x^3-6x^2-10x+10$

Now at this point, I realized that both sides were equal, and that I must have done something very, very wrong ;\. However, I worked through it again and got the same result.

Can someone point out what I'm doing wrong here? This question is driving me nuts.

any help will be much appreciated,

Coukapecker

2. ## Re: Polynomial Division with Variables n and m

Hello, Coukapecker!

When $10x^3+mx^2-x+10$ is divided by $5x-3,$
the quotient is $2x^2+nx-2$ and the remainder is $4.$
Find the values of $m$ and $n.$

The quotient times the divisor plus the remainder should equal the dividend, so:

. . $10x^3+mx^2-x+10\:=\:(5x-3)(2x^2+nx-2)+4$

Expand the right side:
. . $10x^3+mx^2-x+10\:=\:10x^3 + 5nx^2 - 10x - 6x^2 - 3nx + 6 + 4$

. . $10x^3 + mx^2 - x + 10 \:=\:10x^3 + (5n-6)x^2 - (3n+10)x + 10$

Equate coefficients: . $\begin{Bmatrix}10 \:=\:10 & [1] \\ m \:=\:5n-6 & [2] \\ \text{-}1 \:=\:\text{-}(3n+10) & [3] \\ 10 \:=\:10 & [4] \end{Bmatrix}$

From [3]: . $\text{-}1 \:=\:\text{-}3n - 10 \quad\Rightarrow\quad \boxed{n \:=\:\text{-3}}$

Substitute into [2]: . $m \:=\:5(\text{-}3)-6 \quad\Rightarrow\quad \boxed{m \:=\:\text{-}21}$

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### mx divided by x

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