Evaluate: 3x^2-48/3(x+4)(x-4) for x=4

A)0
B)1
C)2
D)undefined

2) Determine the non-permissible value(s) of 3x(x-2)/x+2 *TIMES* x+2/3-x
A)3
B)3,-2
C)2,-3
D)-2

And please explain how you found the answer. Thank you.

I think you meant

(3x^2-48) / [3(x+4)(x-4)]

= [3(x^2 - 16)] / [3(x+4)(x-4)]

= [3(x+4)(x-4)] / [3(x+4)(x-4)]

= 1 -- (b)

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[3x(x-2)/(x+2)] * [(x+2)/(3-x)]

= [3x(x-2)] * [1/(3-x)]

= 3x(x-2)/(3-x)

the non-permissible value is 3 ; i.e., when 3 - x = 0 -- (a)

take 3 comman frm numerator n factorize the remaining after taking 3 comman frm numerator it will give u same as in denominator...hence ans is one