Answer :
The zeros of the function d(x) = 15x^3 - 48x^2 - 48x can be found by factoring out common factors. The zeros are x = 0 with multiplicity 1 and x = 4 with multiplicity 2.
The zeros of the function d(x) = 15x^3 - 48x^2 - 48x, we set the function equal to zero and factor out common terms if possible.
d(x) = 15x^3 - 48x^2 - 48x = 0
Factoring out an x from each term, we have:
x(15x^2 - 48x - 48) = 0
Now, we need to solve the equation by factoring the quadratic expression within the parentheses.
15x^2 - 48x - 48 = 0
Factoring out a common factor of 3, we get:
3(5x^2 - 16x - 16) = 0
Next, we can factor the quadratic expression further:
3(5x + 4)(x - 4) = 0
Setting each factor equal to zero, we find:
5x + 4 = 0 -> x = -4/5
x - 4 = 0 -> x = 4
Therefore, the zeros of the function are x = -4/5 with multiplicity 1 and x = 4 with multiplicity 2.
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