For the given rational functions f(x) and g(x), f(x) -g(x).
-$$f ( x ) = \frac { 3 x ^ { 2 } + 12 } { x ^ { 2 } + 3 x - 4 } , g ( x ) = \frac { 13 x } { x ^ { 2 } + 3 x - 4 }$$
A) $\frac { ( x + 3 ) ( 3 x + 4 ) } { ( x + 4 ) ( x - 1 ) }$
B) $\frac { ( x - 3 ) ( 3 x + 4 ) } { ( x + 4 ) ( x - 1 ) }$
C) $\frac { ( x - 3 ) ( 3 x - 4 ) } { ( x + 4 ) ( x - 1 ) }$
D) $\frac { 3 x - 4 } { x - 1 }$

Question

Quizplus · Accepted Answer

To subtract two rational functions, we need to have a common denominator. In this case, the common denominator is $(x+4)(x-1)$. We then need to adjust each fraction so it has this denominator.

For $f(x)$, we already have this denominator, so we don't need to make any changes:

$$f(x)=\frac{3x^2+12}{x^2+3x-4}=\frac{3(x+4)(x-1)}{(x+4)(x-1)}=\frac{3(x+4)}{x-1}.$$

For $g(x)$, we need to rewrite it with the common denominator:

$$g(x)=\frac{13x}{x^2+3x-4}=\frac{13x}{(x+4)(x-1)}.$$

Combining these fraction with the common denominator, we have

$$f(x)-g(x)=\frac{3(x+4)}{x-1}-\frac{13x}{(x+4)(x-1)}.$$

Multiplying the second term by $\frac{x+4}{x+4}$ in the numerator and simplifying, we get

$$f(x)-g(x)=\frac{3(x+4)(x+4)-13x}{(x+4)(x-1)}=\frac{3x^2+16x-16}{(x+4)(x-1)}.$$

Expanding the numerator and factoring, we get

$$f(x)-g(x)=\frac{(x-3)(3x-4)}{(x+4)(x-1)}.$$

Therefore, the answer is C.

For the Given Rational Functions F(x) and G(x), F(x) -G(x) f(x)=3x2+12x2+3x−4,g(x)=13xx2+3x−4f ( x ) = \frac { 3 x ^ { 2 } + 12 } { x ^ { 2 } + 3 x - 4 } , g ( x ) = \frac { 13 x } { x ^ { 2 } + 3 x - 4 }f(x)=x2+3x−43x2+12​,g(x)=x2+3x−413x​

For the Given Rational Functions F(x) and G(x), F(x) -G(x) $f ( x ) = \frac { 3 x ^ { 2 } + 12 } { x ^ { 2 } + 3 x - 4 } , g ( x ) = \frac { 13 x } { x ^ { 2 } + 3 x - 4 }$