parkhydel wrote:
If \(x ≠ -\frac{1}{2}\), then \(\frac{6x^3 + 3x^2 - 8x - 4}{2x + 1} =\)
A. \(3x^2 + \frac{3}{2}x -8\)
B. \(3x^2 + \frac{3}{2}x -4\)
C. \(3x2 – 4\)
D. \(3x – 4\)
E. \(3x + 4\)
If you're not sure how to simplify the given expression, you can also solve this question by
testing values.
Since we're looking for an
equivalent expression, both the given expression and the correct answer must evaluate to have the
same value for any value of x.
So, for example, if \(x = 2\), then \(\frac{6x^3 + 3x^2 - 8x - 4}{2x + 1} =\frac{6(2)^3 + 3(2)^2 - 8(2) - 4}{2(2) + 1}\)
\(=\frac{48 + 12 - 16 - 4}{4 + 1}\)
\(=\frac{40}{5}\)
\(=8\)
This tells us that, the correct answer must also evaluate to be \(8\) when \(x = 2\)
So let's plug \(x = 2\) into each answer choice and see which expression evaluates to be \(8\)
A. \(3(2)^2 + \frac{3}{2}(2) -8=7\). No good. We need the expression to evaluate to be \(8\)
B. \(3(2)^2 + \frac{3}{2}(2) -4=11\). No good. We need the expression to evaluate to be \(8\)
C. \(3(2^2) – 4=8\). Woo woo!!!
D. \(3(2) – 4=2\). No good. We need the expression to evaluate to be \(8\)
E. \(3(2) + 4=10\). No good. We need the expression to evaluate to be \(8\)
Answer: C
Cheers,
Brent
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