show your work. find the product. $\frac{3m^{3}}{m^{2}-25}cdot\frac{m^{2}+6m + 5}{m^{2}}$

Explanation:

Step1: Factor the expressions

Factor $m^{2}-25$ using the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$. Here, $a = m$ and $b = 5$, so $m^{2}-25=(m + 5)(m - 5)$. Factor $m^{2}+6m + 5$ using the formula $x^{2}+(a + b)x+ab=(x + a)(x + b)$. Here, $a = 1$ and $b = 5$, so $m^{2}+6m + 5=(m + 1)(m + 5)$. The expression becomes $\frac{3m^{3}}{(m + 5)(m - 5)}\cdot\frac{(m + 1)(m + 5)}{m^{2}}$.

Step2: Cancel out the common factors

Cancel out the common factors $m^{2}$ and $(m + 5)$ in the numerator and the denominator. $\frac{3m^{3}}{(m + 5)(m - 5)}\cdot\frac{(m + 1)(m + 5)}{m^{2}}=\frac{3m\cdot(m + 1)}{m - 5}=\frac{3m^{2}+3m}{m - 5}$

Answer:

$\frac{3m^{2}+3m}{m - 5}$