For the given functions $f(x)$ and $g(x)$, find $f(x) \cdot g(x)$.
-$f(x)=x+8, g(x)=\frac{x^{2}+5 x-24}{4 x^{2}+24 x-64}$

A) $\frac{x-3}{4 x-8}$

B) $\frac{(x+8)(x-3)}{4 x-8}$

C) $\frac{(x+8)^{2}(x-3)}{4 x-8}$

D) $\frac{(x-8)(x-3)}{4 x+8}$

Question

For the given functions $f(x)$ and $g(x)$, find $f(x) \cdot g(x)$.
-$f(x)=x+8, g(x)=\frac{x^{2}+5 x-24}{4 x^{2}+24 x-64}$

A) $\frac{x-3}{4 x-8}$

B) $\frac{(x+8)(x-3)}{4 x-8}$

C) $\frac{(x+8)^{2}(x-3)}{4 x-8}$

D) $\frac{(x-8)(x-3)}{4 x+8}$

Quizplus · Accepted Answer

To find $f(x) \cdot g(x)$, we multiply $f(x) = x + 8$ by $g(x) = \frac{x^{2} + 5x - 24}{4x^{2} + 24x - 64}$. Simplifying the denominator of $g(x)$ by factoring, we get $4x^{2} + 24x - 64 = 4(x^{2} + 6x - 16) = 4(x + 8)(x - 2)$. The numerator of $g(x)$ can be factored as $(x + 8)(x - 3)$. Thus, $g(x) = \frac{(x + 8)(x - 3)}{4(x + 8)(x - 2)}$. When we multiply $f(x)$ and $g(x)$, the $(x + 8)$ in $f(x)$ and one $(x + 8)$ in $g(x)$ cancel out with the $(x + 8)$ in the denominator of $g(x)$, leaving us with $\frac{(x + 8)(x - 3)}{4(x - 2)}$, which simplifies to option B.

For the Given Functions f(x)f(x)f(x) And g(x)g(x)g(x) , Find f(x)⋅g(x)f(x) \cdot g(x)f(x)⋅g(x) -

For the Given Functions $f(x)$ And $g(x)$ , Find $f(x) \cdot g(x)$
-