Question

Question #d40e0

Answer 1

`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 = 5n-6" [1]"`

Matching the coefficient of the `x^2` term on the left with the coefficient of the `x^2` term on the right:

`-1 = -10-3n" [2]"`

Use equation [2] to solve for n:

`1 = 10+3n`

`3n = -9`

`n = -3`

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

`m = 5(-3)-6`

`m = -21`