Add.
-$$\frac { x + 2 } { x ^ { 2 } + 12 x + 35 } + \frac { 3 x - 5 } { x ^ { 2 } + 9 x + 20 }$$
A) $4 x - 3$
B) $\frac { 4 x - 3 } { 2 x ^ { 2 } + 21 x + 55 }$
C) $\frac { 4 x ^ { 2 } + 22 x - 27 } { ( x + 5 ) ( x + 7 ) ( x + 4 ) }$
D) $\frac { 4 x ^ { 2 } + 22 x - 27 } { ( x - 5 ) ( x - 7 ) ( x - 4 ) }$

Question

Quizplus · Accepted Answer

We need to factor both denominators of the fractions appearing in the expression. 
$$x^2+12x+35=(x+7)(x+5)$$
$$x^2+9x+20=(x+5)(x+4)$$
Thus, the given expression can be written as 
$$\frac{x+2}{(x+7)(x+5)}+\frac{3x-5}{(x+5)(x+4)}=\frac{(x+2)(x+4)+(3x-5)(x+7)}{(x+7)(x+5)(x+4)}$$
On simplifying the numerator, we get
\begin{align*}
&(x+2)(x+4)+(3x-5)(x+7)\
&=x^2+6x+8+3x^2+16x-35\
&=4x^2+22x-27
\end{align*}
Therefore, the final answer is 
$$\frac{4x^2+22x-27}{(x+7)(x+5)(x+4)}$$
which matches with choice (C).

Add - x+2x2+12x+35+3x−5x2+9x+20\frac { x + 2 } { x ^ { 2 } + 12 x + 35 } + \frac { 3 x - 5 } { x ^ { 2 } + 9 x + 20 }x2+12x+35x+2​+x2+9x+203x−5​

Add
- $\frac { x + 2 } { x ^ { 2 } + 12 x + 35 } + \frac { 3 x - 5 } { x ^ { 2 } + 9 x + 20 }$