For the given rational functions f(x) and g(x), f(x) -g(x).
-$$f ( x ) = \frac { 2 x + 6 } { x ^ { 2 } + 4 x + 3 } , g ( x ) = \frac { x + 3 } { x ^ { 2 } + 4 x + 3 }$$
A) $\frac { 1 } { x + 1 }$
B) $\frac { 7 } { x ^ { 2 } + 4 x + 3 }$
C) $x + 3$
D) $\frac { x + 9 } { x ^ { 2 } + 4 x + 3 }$

Question

Quizplus · Accepted Answer

To find the difference of two rational functions, we need to subtract the numerators of the two fractions while keeping the same denominator. Therefore, 
$$f(x)-g(x)=\frac{2x+6}{x^2+4x+3}-\frac{x+3}{x^2+4x+3}=\frac{(2x+6)-(x+3)}{x^2+4x+3}=\frac{x+3}{x^2+4x+3}$$
We can simplify this expression by factoring the denominator and cancelling the common factor in the numerator and the denominator:
$$\frac{x+3}{x^2+4x+3}=\frac{x+3}{(x+1)(x+3)}=\frac{1}{x+1}$$
Therefore, the answer is choice A.

For the Given Rational Functions F(x) and G(x), F(x) -G(x) f(x)=2x+6x2+4x+3,g(x)=x+3x2+4x+3f ( x ) = \frac { 2 x + 6 } { x ^ { 2 } + 4 x + 3 } , g ( x ) = \frac { x + 3 } { x ^ { 2 } + 4 x + 3 }f(x)=x2+4x+32x+6​,g(x)=x2+4x+3x+3​

For the Given Rational Functions F(x) and G(x), F(x) -G(x) $f ( x ) = \frac { 2 x + 6 } { x ^ { 2 } + 4 x + 3 } , g ( x ) = \frac { x + 3 } { x ^ { 2 } + 4 x + 3 }$