Bunuel wrote:
If x ≠ –2, then \(\frac{7x^2 + 28x + 28}{(x + 2)^2} =\)
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11
Another approach:
Plug in a nice value for x and evaluate the expression.
A super nice x-value is x = 0
We get:\(\frac{7(0^2) + 28(0) + 28}{(0 + 2)^2} = \frac{28}{2^2}\)
\(= \frac{28}{4}\)
\(= 7\)
Answer:
[Reveal] Spoiler:
still A ![Image]()
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