A(t)=(t−k)(t−3)(t−6)(t+3) is a polynomial function of tt, where kk is a constant. given that a(2)=0a(2)=0, what is the absolute value of the product of the zeros of aa?

The absolute value of the product of the zeros of a(t) is 108.

Step by step explanation:

The given polynomial function is

Where k is the constant.

To find the zeros of a(t), equate the polynomial equal to zero.

By using zero product property, equate each factor equal to zero.

Therefore the zeros of the function are k, 3, 6 and -3. Since it is given that

Therefore, 2 is a zero of a(t). So, the value of k is 2.

The product of zeros is

The absolute value of the product is 108.

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