Question

$\frac{5}{3}-\frac{m + 6}{3m^{2}-18m + 27}=\frac{am^{2}+bm + c}{3(m - 3)^{2}}
a=

b=

c=$

Explanation:

Step1: Factor the denominator

First, factor $3m^{2}-18m + 27=3(m^{2}-6m + 9)=3(m - 3)^{2}$.

Step2: Find a common - denominator

The first fraction $\frac{5}{3}$ needs to be rewritten with the denominator $3(m - 3)^{2}$. So, $\frac{5}{3}=\frac{5(m - 3)^{2}}{3(m - 3)^{2}}=\frac{5(m^{2}-6m + 9)}{3(m - 3)^{2}}=\frac{5m^{2}-30m + 45}{3(m - 3)^{2}}$, and the second fraction is $\frac{m + 6}{3(m - 3)^{2}}$.

Step3: Subtract the fractions

$\frac{5m^{2}-30m + 45-(m + 6)}{3(m - 3)^{2}}=\frac{5m^{2}-30m + 45 - m-6}{3(m - 3)^{2}}=\frac{5m^{2}-31m + 39}{3(m - 3)^{2}}$.

Answer:

A = 5
B = - 31
C = 39