Question

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? Answer:

Answer 1

Answer:

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

Step by step explanation:

The given polynomial function is

$a(t)=(t-k)(t-3)(t-6)(t+3)$

Where k is the constant.

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

$(t-k)(t-3)(t-6)(t+3)=0$

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

$t-k=0$

$t=k$

$t-3=0$

$t=3$

$t-6=0$

$t=6$

$t+3=0$

$t=-3$

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

$a(2)=0$

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

The product of zeros is

$2\times 3\times 6\times -3=-108$

The absolute value of the product is 108.