### Instruction: Find the range of the function
\[h(x) = \frac{2(x + 7)(x - 3)}{x + 7}.\].### Response: To find the range of a function, I need to consider what possible values it can output for different inputs.
I notice that the function has a factor of x + 7 in both the numerator and the denominator.
This means I can simplify the function by canceling out this factor, as long as x is not equal to -7.
So, for x not equal to -7, I have
\[h(x) = \frac{2(x + 7)(x - 3)}{x + 7} = 2(x - 3).\]
This is a linear function with a slope of 2 and a y-intercept of -6.
I know that linear functions have a range of all real numbers, since they can output any value as x changes.
So, for x not equal to -7, the range of h(x) is all real numbers.
But what happens when x is equal to -7?
In that case, the original function is undefined, since it would have a zero in the denominator.
This means that h(-7) does not exist, and the function has a hole at x = -7.
This hole means that there is one value that the function cannot output, and that is the y-value of the hole.
To find this y-value, I plug in x = -7 into the simplified function, and get
\[h(-7) = 2(-7 - 3) = -20.\]
So, the function h(x) cannot output -20, even though it can output any other real number.
Therefore, the range of h(x) is all real numbers except -20.