To determine which function has exactly three distinct real zeros, let's analyze each given function and find its distinct real zeros.
### Function A: [tex]\( h(x) = (x-9)^2 (x-4)^2 \)[/tex]
This is a product of two squared terms:
[tex]\( (x-9)^2 \)[/tex] which has a double zero at [tex]\( x = 9 \)[/tex]
[tex]\( (x-4)^2 \)[/tex] which has a double zero at [tex]\( x = 4 \)[/tex]
Therefore, the distinct real zeros are [tex]\( x = 9 \)[/tex] and [tex]\( x = 4 \)[/tex]. So, there are 2 distinct real zeros.
### Function B: [tex]\( h(x) = x(x+7)^2 \)[/tex]
This function can be broken down into:
[tex]\( x \)[/tex] which has a zero at [tex]\( x = 0 \)[/tex]
[tex]\( (x+7)^2 \)[/tex] which has a double zero at [tex]\( x = -7 \)[/tex]
Therefore, the distinct real zeros are [tex]\( x = 0 \)[/tex] and [tex]\( x = -7 \)[/tex]. So, there are 2 distinct real zeros.
### Function C: [tex]\( h(x) = (x-3)(x+1)(x+3)(x+8) \)[/tex]
Each factor will give exactly one zero:
[tex]\( (x-3) \)[/tex] which has a zero at [tex]\( x = 3 \)[/tex]
[tex]\( (x+1) \)[/tex] which has a zero at [tex]\( x = -1 \)[/tex]
[tex]\( (x+3) \)[/tex] which has a zero at [tex]\( x = 0 \)[/tex]
[tex]\( (x+8) \)[/tex] which has a zero at [tex]\( x = -8 \)[/tex]
Therefore, the distinct real zeros are [tex]\( x = 3 \)[/tex], [tex]\( x = -1 \)[/tex], [tex]\( x = 3 \)[/tex], and [tex]\( x = -8 \)[/tex]. So, there are 4 distinct real zeros.
### Function D: [tex]\( h(x) = (x-2)^2(x+4)(x-1) \)[/tex]
This function can be broken down into:
[tex]\( (x-2)^2 \)[/tex] which has a double zero at [tex]\( x = 2 \)[/tex]
[tex]\( (x+4) \)[/tex] which has a zero at [tex]\( x = -4 \)[/tex]
[tex]\( (x-1) \)[/tex] which has a zero at [tex]\( x = 1 \)[/tex]
Therefore, the distinct real zeros are [tex]\( x = 2 \)[/tex], [tex]\( x = -4 \)[/tex], and [tex]\( x = 1 \)[/tex]. So, there are 3 distinct real zeros.
Given these analyses, the function that has exactly three distinct real zeros is:
D. [tex]\(h(x) = (x-2)^2(x+4)(x-1) \)[/tex]
