Question

fa25 math 125 - 09: precalculus i homework: ch 1.2 operations with algebraic expressions question 14, ar.3.82 divide. $\frac{9m^{3}+18m^{2}+15m}{3m^{2}}$ (simplify your answer.)

Explanation:

Step1: Split the fraction

$\frac{9m^{3}+18m^{2}+15m}{3m^{2}}=\frac{9m^{3}}{3m^{2}}+\frac{18m^{2}}{3m^{2}}+\frac{15m}{3m^{2}}$

Step2: Simplify each term

For $\frac{9m^{3}}{3m^{2}}$, using the rule $\frac{a^{m}}{a^{n}}=a^{m - n}$, we have $\frac{9}{3}m^{3-2}=3m$.
For $\frac{18m^{2}}{3m^{2}}$, $\frac{18}{3}=6$.
For $\frac{15m}{3m^{2}}$, $\frac{15}{3}\times\frac{m}{m^{2}} = 5\times m^{1 - 2}= \frac{5}{m}$.

Answer:

$3m + 6+\frac{5}{m}$