Question

a(t)=(t-k)(t-3)(t-6)(t+3) is a polynomial function of t, where k is a constant. given that a(2)=0, what is the absolute value of the product of the zeroes of a?

Answer 1

$\begin{gathered}{\Huge{\textsf{\textbf{\underline{\underline{\purple{Answer:}}}}}}}\end{gathered}$

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.

tk) (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,

a(2) = 0

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

The product of zeros is

2 × 3 ×6×-3 = -108

The absolute value of the product is 108.

• Please Drop Some Thanks.

Answer 2

put the value of t=2 and we get (2-k)20=0

therefore, k=2