Question

$\frac{5}{2v^{3}}+\frac{3v}{v^{2}-v - 20}$
$\frac{av^{4}+bv^{2}+cv + d}{2v^{3}(v - 5)(v - 4)}$
$a=$
$b=$
$c=$
$d=$

Explanation:

Step1: Factor the denominator

Factor $v^{2}-v - 20=(v - 5)(v+4)$.

Step2: Find a common denominator

The common denominator of $\frac{5}{2v^{3}}$ and $\frac{3v}{v^{2}-v - 20}$ is $2v^{3}(v - 5)(v + 4)$.
Rewrite $\frac{5}{2v^{3}}=\frac{5(v - 5)(v + 4)}{2v^{3}(v - 5)(v + 4)}=\frac{5(v^{2}-v - 20)}{2v^{3}(v - 5)(v + 4)}=\frac{5v^{2}-5v - 100}{2v^{3}(v - 5)(v + 4)}$ and $\frac{3v}{v^{2}-v - 20}=\frac{3v\times2v^{3}}{2v^{3}(v - 5)(v + 4)}=\frac{6v^{4}}{2v^{3}(v - 5)(v + 4)}$.

Step3: Add the fractions

$\frac{5}{2v^{3}}+\frac{3v}{v^{2}-v - 20}=\frac{6v^{4}+5v^{2}-5v - 100}{2v^{3}(v - 5)(v + 4)}$.

Answer:

$A = 6$
$B = 5$
$C=-5$
$D=-100$