Question

$\frac{6}{3v^{2}-21v + 36}-\frac{6}{3v}$
$\frac{av^{2}+bv + c}{v(v - 4)(v - 3)}$
$a=$
$b=$
$c=$

Explanation:

Step1: Factor the denominator of the first - fraction

First, factor out 3 from $3v^{2}-21v + 36$: $3v^{2}-21v + 36=3(v^{2}-7v + 12)=3(v - 3)(v - 4)$. So, $\frac{6}{3v^{2}-21v + 36}=\frac{6}{3(v - 3)(v - 4)}=\frac{2}{(v - 3)(v - 4)}$. And $\frac{6}{3v}=\frac{2}{v}$.

Step2: Find a common denominator

The common denominator of $\frac{2}{(v - 3)(v - 4)}$ and $\frac{2}{v}$ is $v(v - 3)(v - 4)$. Rewrite the fractions: $\frac{2}{(v - 3)(v - 4)}\times\frac{v}{v}=\frac{2v}{v(v - 3)(v - 4)}$ and $\frac{2}{v}\times\frac{(v - 3)(v - 4)}{(v - 3)(v - 4)}=\frac{2(v^{2}-7v + 12)}{v(v - 3)(v - 4)}=\frac{2v^{2}-14v + 24}{v(v - 3)(v - 4)}$.

Step3: Subtract the fractions

$\frac{2v}{v(v - 3)(v - 4)}-\frac{2v^{2}-14v + 24}{v(v - 3)(v - 4)}=\frac{2v-(2v^{2}-14v + 24)}{v(v - 3)(v - 4)}=\frac{-2v^{2}+16v - 24}{v(v - 3)(v - 4)}$. Comparing with $\frac{Av^{2}+Bv + C}{v(v - 3)(v - 4)}$, we have $A=-2$, $B = 16$, $C=-24$.

Answer:

$A=-2$
$B = 16$
$C=-24$