Answer:
OPTION A: [tex]$ \frac{12}{1 + 8x} $[/tex], where [tex]$ x \ne - \frac{1}{8} $[/tex].
Step-by-step explanation:
Given: [tex]$ \frac{- 12x^4}{x^4 + 8x^5} $[/tex]
Taking [tex]$ x^4 $[/tex] common outside in the denominator, we get:
[tex]$ = \frac{-12x^4}{(x^4)(1 + 8x)} $[/tex]
[tex]$ x^4 $[/tex] will get cancelled on the numerator and denominator, we get:
[tex]$ = \frac{-12}{1 + 8x} $[/tex]
we know that the denominator can not be zero.
That means, 1 + 8x [tex]$ \ne $[/tex] 0.
[tex]$ \implies 8x \ne -1 $[/tex]
[tex]$ \implies x \ne \frac{-1}{8} $[/tex]
So, the answer is: [tex]$ \frac{-12}{1 + 8x} $[/tex], where [tex]$ x \ne \frac{-1}{8} $[/tex].
