Answer
$2x^2-x-5+\frac{3}{x-5}$
Work Step by Step
Step 1. Rewrite and factor the numerator as $2x^3-11x^2+28=2x^3-10x^2-x^2+25+3=2x^2(x-5)-(x^2-25)+3=2x^2(x-5)-(x+5)(x-5)+3$
Step 2. Thus, we have $\frac{2x^3-11x^2+28}{x-5}=\frac{2x^2(x-5)-(x+5)(x-5)+3}{x-5}=2x^2-(x+5)+\frac{3}{x-5}=2x^2-x-5+\frac{3}{x-5}$
