Question

$\frac{x + 6}{x^{2}+x - 20}+\frac{4}{5}
\frac{ax^{2}+bx + c}{5(x - 4)(x + 5)}
a=

b=

c=$

Explanation:

Step1: Factor the denominator

Factor $x^{2}+x - 20$ as $(x - 4)(x+5)$. So the first - fraction is $\frac{x + 6}{(x - 4)(x + 5)}$.

Step2: Find a common denominator

The common denominator of $\frac{x + 6}{(x - 4)(x + 5)}$ and $\frac{4}{5}$ is $5(x - 4)(x + 5)$. Rewrite the fractions: $\frac{x + 6}{(x - 4)(x + 5)}\times\frac{5}{5}=\frac{5(x + 6)}{5(x - 4)(x + 5)}$ and $\frac{4}{5}\times\frac{(x - 4)(x + 5)}{(x - 4)(x + 5)}=\frac{4(x^{2}+x - 20)}{5(x - 4)(x + 5)}$.

Step3: Add the fractions

$\frac{5(x + 6)+4(x^{2}+x - 20)}{5(x - 4)(x + 5)}=\frac{5x+30 + 4x^{2}+4x-80}{5(x - 4)(x + 5)}=\frac{4x^{2}+9x - 50}{5(x - 4)(x + 5)}$.

Answer:

$A = 4$
$B = 9$
$C=-50$