Question #d40e0

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#