Question

Which is one of the binomial factors of the polynomial x^3+3x-2x-8?

Answer 1

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.

Answer 2

Answer:

x+2

Step-by-step explanation:

A factor of a polynomial can be thought of as the value of x at which the polynomial is equal to zero.

So, you can use the values in the options given and put them in the polynomial from the question.

for example: try, part a) x-1, here x=1 since the factor is x-1=0

put this value in the polynomial to see if it results to zero.

$(1)^3 + 3(1)^2 - 2(1) - 8\\ -6$

so this isn't the answer.

now try, part d) x + 2, here, x = -2

$(-2)^3 + 3(-2)^2 - 2(-2) - 8\\ 0$

you'll see this is the factor!