To find other zeros of the polynomial [tex]\( f(x) = 2x^3 + 9x^2 - 8x - 36 \)[/tex] when [tex]\( x = -2 \)[/tex] is already known to be a zero, we follow these steps:
### Step 1: Verify [tex]\( x = -2 \)[/tex] is a zero
First, confirm that [tex]\( x = -2 \)[/tex] is indeed a zero by substituting [tex]\( x = -2 \)[/tex] into the polynomial:
[tex]\[ f(-2) = 2(-2)^3 + 9(-2)^2 - 8(-2) - 36 \][/tex]
[tex]\[ f(-2) = 2(-8) + 9(4) + 16 - 36 \][/tex]
[tex]\[ f(-2) = -16 + 36 + 16 - 36 \][/tex]
[tex]\[ f(-2) = 0 \][/tex]
Since [tex]\( f(-2) = 0 \)[/tex], [tex]\( x = -2 \)[/tex] is confirmed as a zero.
### Step 2: Polynomial factorization
Given one zero [tex]\( x = -2 \)[/tex], we can factorize the polynomial [tex]\( f(x) \)[/tex]. We know that [tex]\( (x + 2) \)[/tex] is a factor of [tex]\( f(x) \)[/tex]. We perform polynomial division or use factoring techniques to find the complete factorization of [tex]\( f(x) \)[/tex].
### Step 3: Use known factorization results
The polynomial can be factored into:
[tex]\[ f(x) = 2(x + 2)(x + \frac{9}{2})(x - 2) \][/tex]
This means that the zeros of [tex]\( f(x) \)[/tex] are:
[tex]\[ x = -2, \ \ x = -\frac{9}{2}, \ \ x = 2 \][/tex]
### Step 4: Identify other zeros
From the factorization, aside from the zero [tex]\( x = -2 \)[/tex], the other zeros are [tex]\( x = -\frac{9}{2} \)[/tex] and [tex]\( x = 2 \)[/tex].
Among the zero points given in the problem:
[tex]\[ x = 8 \][/tex]
[tex]\[ x = 4 \][/tex]
[tex]\[ x = 3 \][/tex]
[tex]\[ x = 2 \][/tex]
The option [tex]\( x = 2 \)[/tex] is listed among these, which matches the zero we have identified earlier.
Therefore, another zero of [tex]\( f(x) \)[/tex] is:
[tex]\[ x = 2 \][/tex]
Thus, the correct answer is:
[tex]\[ x = 2 \][/tex]
