Question

Find the third degree polynomial function that has an output of 40 when x=1, and has zeros −19 and −i?

Answer 1

`f(x) = (x + 19)(x^2 + 1)`

Explanation:

Complex numbers:

The following relation is important for complex numbers:

`i^2 = -1`

Zeros of a function:

Given a polynomial f(x), this polynomial has roots
`x_(1), x_(2), x_(n)` such that it can be written as:
`a(x - x_(1))*(x - x_(2))*...*(x-x_n)`, in which a is the leading coefficient.

Has zeros −19 and −i

If -i is a zero, its conjugate i is also a zero. So

`f(x) = a(x - (-19))(x - (-i))(x - i) = a(x+19)(x+i)(x-i) = a(x+19)(x^2 - i^2) = a(x + 19)(x^2 + 1)`

Output of 40 when x=1

This means that when
`x = 1, f(x) = 40`. We use this to find the leading coefficient a. So

`f(x) = a(x + 19)(x^2 + 1)`

`40 = a(20)(2)`

`40a = 40`

`a = 1`

The polynomial is:

`f(x) = (x + 19)(x^2 + 1)`