Question

Construct a polynomial function with the following properties: third degree, only real coefficients, −3 and 3+i are two of the zeros, y-intercept is −90.

Answer 1

Answer:

$\boxed{-3(x+3)(x^2-6x+10)}$

Step-by-step explanation:

Hello,

As the polynomial has only real coefficients, it means that 3-i is a zero too, because we apply the Conjugate Zeros Theorem.

It means that we can write the expression as below, k being a real number that we will have to identify.

$k(x+3)(x-3-i)(x-3+i)=k(x+3)((x-3)^2-i^2)\\\\=k(x+3)(x^2-6x+9+1)\\\\=k(x+3)(x^2-6x+10)$

And for x = 0, y = -90 so we can write

-90=k*3*10, meaning that k=-3

Hope this helps.

Do not hesitate if you need further explanation.

Thank you