# Question d40e0

May 3, 2017

$n = - 3$ and $m = - 21$

#### Explanation:

Given:

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

Please observe that we have written the remainder over the divisor; this is a way to write the quotient and include the remainder without setting it apart.

Multiply both sides by the divisor:

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

The multiplying by divisor cancels the denominator of the remainder

10x^3+mx^2−x+10 = (5x-3)(2x^2+nx−2)+4

Split the binomial into two parts:

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

Distribute the monomials across their respective trinomial:

10x^3+mx^2−x+10 = 10x^3+5nx^2−10x-6x^2-3nx+6+4

On the right, group the coefficients of the ${x}^{2}$ terms and the x terms, respectively:

10x^3+mx^2−x+10 = 10x^3+(5n-6)x^2+(-10-3n)x+10#

Matching the coefficient of the ${x}^{2}$ term on the left with the coefficient of the ${x}^{2}$ term on the right:

$m = 5 n - 6 \text{ [1]}$

Matching the coefficient of the ${x}^{2}$ term on the left with the coefficient of the ${x}^{2}$ term on the right:

$- 1 = - 10 - 3 n \text{ [2]}$

Use equation [2] to solve for n:

$1 = 10 + 3 n$

$3 n = - 9$

$n = - 3$

Substitute -3 for n into equation [1] and solve for m:

$m = 5 \left(- 3\right) - 6$

$m = - 21$