As we see, there are only variables in the question and in the answer choices. We can TEST some values and check the options.
Let's assume x = 6 as the denominator will be positive.
Substituting x = 6 in the equation we get
\([ \frac{((6^3) - 6(6^2) - 6 + 30) }{ (6-5)} ]\)
= \([ \frac{( - 6 + 30) }{ 1 } ]\)
= 24
Now, let's check the options.
A. \((x^2) - x - 6 = (6^2) - 6 - 6 = 36 - 12 = 24\). This matches, but let's check other options as well.
B. \((x^3) + 3(x^2) + 3x = (6^3) + 3(6^2) + 3 \times 6\). everything is added, will result in a very big number. Eliminate
C. \((x^3) - 25 = (6^3) - 25 = 216 - 25\). No need to calculate. Very big than 24. Eliminate
D. \((x^3) - 5(x^2) - 3x = (6^3) - 5(6^2) - 3 \times 6 = 216 - 5 \times 36 - 18 = 216 - 180 - 18 = 36 - 18 = 18\). Eliminate.
E. \((x^2) + x + 10 = (6^2) + 6 + 10\). All terms added, will be greater than 24. Eliminate.
Hence, our option becomes A.
OA, Aanon321 wrote:
If H = [(x^3) - 6(x^2) - x + 30] / (x-5)and x ≠ 5, then H is equivalent to which of the following?
A. (x^2) - x - 6
B. (x^3) + 3(x^2) + 3x
C. (x^3) - 25
D. (x^3) - 5(x^2) - 3x
E. (x^2) + x + 10
I know the problem can be solved by assuming x=2 and then checking which equation yields the correct answer (the correct answer is option A), but I want to understand how you would solve the equation through algebra as well.
Thanks!