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:

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

Expand the right side:

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

Cancel Terms:

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

Factor:

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

Cancel:

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

Isolate m:

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

Sub m into initial equation to solve for n:

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

Expand:

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

Collect Like Terms:

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

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