For the given rational functions $f(x)$ and $g(x), f(x)$ - $g(x)$.
-$f(x)=\frac{9 x-2}{x^{2}+5 x-14}, g(x)=\frac{x+4}{x^{2}-4}$

A) $\frac{10 x+2}{(x-2)(x+7)}$

B) $\frac{8 \mathrm{x}-6}{(\mathrm{x}-2)(\mathrm{x}+2)}$

C) $\frac{8 x^{2}+5 x-32}{(x-2)(x+2)(x+7)}$

D) $\frac{10 x^{2}+27 x+24}{(x-2)(x+2)(x+7)}$

Question

For the given rational functions $f(x)$ and $g(x), f(x)$ - $g(x)$.
-$f(x)=\frac{9 x-2}{x^{2}+5 x-14}, g(x)=\frac{x+4}{x^{2}-4}$

A) $\frac{10 x+2}{(x-2)(x+7)}$

B) $\frac{8 \mathrm{x}-6}{(\mathrm{x}-2)(\mathrm{x}+2)}$

C) $\frac{8 x^{2}+5 x-32}{(x-2)(x+2)(x+7)}$

D) $\frac{10 x^{2}+27 x+24}{(x-2)(x+2)(x+7)}$

Quizplus · Accepted Answer

To find $f(x) - g(x)$, we first need to express both functions with a common denominator. The denominators of $f(x)$ and $g(x)$ are $x^2 + 5x - 14$ and $x^2 - 4$, which factor to $(x-2)(x+7)$ and $(x-2)(x+2)$, respectively. The least common denominator (LCD) is $(x-2)(x+2)(x+7)$. Adjusting the numerators accordingly and subtracting gives $\frac{8 x^{2}+5 x-32}{(x-2)(x+2)(x+7)}$.

For the Given Rational Functions f(x)f(x)f(x) And g(x),f(x)g(x), f(x)g(x),f(x) - g(x)g(x)g(x) - f(x)=9x−2x2+5x−14,g(x)=x+4x2−4f(x)=\frac{9 x-2}{x^{2}+5 x-14}, g(x)=\frac{x+4}{x^{2}-4}f(x)=x2+5x−149x−2​,g(x)=x2−4x+4​

For the Given Rational Functions $f(x)$ And $g(x), f(x)$ - $g(x)$
- $f(x)=\frac{9 x-2}{x^{2}+5 x-14}, g(x)=\frac{x+4}{x^{2}-4}$