Answer
$3b-8+\frac{2}{b^2+4}$
Work Step by Step
Step 1. Rewrite and factor the numerator as $3b^3-8b^2+12b-30=3b^3+12b-8b^2-32+2=3b(b^2+4)-8(b^2+4)+2=(b^2+4)(3b-8)+2$
Step 2. We have $\frac{3b^3-8b^2+12b-30}{b^2+4}=\frac{(b^2+4)(3b-8)+2}{b^2+4}=3b-8+\frac{2}{b^2+4}$
