Answer:
d. x+2
Step-by-step explanation:
The question is essentially asking which of -2, -1, 1, 2 is a zero of the polynomial. All of them are plausible, because all are factors of -8, the constant term.
So, we don't have much choice but to try them. That means we evaluate the function to see if any of these values of x make it be zero.
±1:
The value 1 is easy to substitute for x, as it makes all of the x-terms equal to their coefficient. Essentially, you add all of the coefficients. Doing that gives ...
1 +3 -2 -8 = -6
Similarly, the value -1 is easy to substitute for x, as it makes all odd-degree terms equal to the opposite of their coefficient. Here, ...
f(-1) = -1 +3 -(-2) -8 = -4
Neither one of these values (-1, +1) is a zero of the polynomial, so choices A and B are eliminated.
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(x-2):
To see if this is a factor, we need to see if x=2 is a zero. Evaluation of a polynomial is sometimes easier when it is written in Horner form:
((x +3)x -2)x -8
Substituting x=2, we get ...
((2 +3)2 -2)2 -8 = (8)2 -8 = 8 . . . not zero
This tells us there is a zero between x=1 and x=2, but that is not what the question is asking.
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(x+2):
We can similarly evaluate the function for x=-2 to see if (x+2) is a factor.
((-2 +3)(-2) -2)(-2) -8 = (-4)(-2) -8 = 0
Since x=-2 makes the function zero, and it makes the factor (x+2) equal to zero, (x+2) is a factor of the polynomial.
So, the factor (x+2) is a factor of the given polynomial.
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I find that a graphing calculator answers questions like this quickly and easily. If you're allowed one, it is a handy tool.
