Question

Is the binomial a factor of the polynomial function?
f(x)=x^3+4x^2−9x−36
Select Yes or No for each binomial.
(x-1)
(x-3)
(x+3)
(x-4)
(x+4)

Answer 1

Answers:

• (x-1) .... no
• (x-3) ... yes
• (x+3) ... yes
• (x-4) .... no
• (x+4) ... yes

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Explanation:

We'll use the rule that if p(k) = 0, then x-k is a factor of p(x). This is the special case of the remainder theorem. Some books may call it the factor theorem.

The term x-1 is in the form x-k with k = 1.

Plug this into the function

$f(x) = x^3+4x^2-9x-36\\f(1) = (1)^3+4(1)^2-9(1)-36\\f(1) = -40$

We end up with a nonzero value, which means (x-1) is not a factor of f(x).

Now let's move to x-3 which shows k = 3

$f(x) = x^3+4x^2-9x-36\\f(3) = (3)^3+4(3)^2-9(3)-36\\f(3) = 0$

We get 0 so this shows (x-3) is a factor of f(x).

Repeat those steps for the following k values

• -3
• 4
• -4

They correspond to (x+3), (x-4), (x+4) in that order.

You should find that the k values -3 and -4 will lead to f(x) = 0. This means (x+3) and (x+4) are also factors of f(x). The input x = 4 leads to f(x) = 56 which means we rule out (x-4).

In summary, (x-3), (x+3) and (x+4) are factors of f(x).

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Another alternative is to graph f(x) as shown below. The x intercepts or roots of f(x) are -4, -3 and 3. Those roots lead to the factors (x+4), (x+3) and (x-3) in that order.