Find all zeros. one zero had been give

The three zeros of the original function f(x) are {-1/2, -3, -5}.

Step-by-step 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}.

he kicked it up? is this the answer ?

step-by-step explanation:

the answer is 11.2 (i know i’m late)