Answer:
The three zeros of the original function f(x) are {-1/2, -3, -5}.
Explanation:
"Synthetic division" is the perfect tool for approaching this problem. Long div. would also "work."
Use -5 as the first divisor in synthetic division:
------------------------
-5 2 17 38 15
-10 -35 -15
--------------------------
2 7 3 0
Note that there's no remainder here. That tells us that -5 is indeed a zero of the given function. We can apply synthetic div. again to the remaining three coefficients, as follows:
-------------
-3 2 7 3
-6 -3
-----------------
2 1 0
Note that the '3' in 2 7 3 tells me that -3, 3, -1 or 1 may be an additional zero. As luck would have it, using -3 as a divisor (see above) results in no remainder, confirming that -3 is the second zero of the original function.
That leaves the coefficients 2 1. This corresponds to 2x + 1 = 0, which is easily solved for x:
If 2x + 1 = 0, then 2x = -1, and x = -1/2.
Thus, the three zeros of the original function f(x) are {-1/2, -3, -5}.