Question

Given a polynomial that has zeros of −5, 4i, and −4i and has a value of 40 when x=3 write the polynomial equation

Answer 1

`f(x) = 0.2(x^3 + 5x^2 + 16x + 80)`

Explanation:

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.

Zeros of −5, 4i, and −4i

This means that
`x_1 = -5, x_2 = 4i, x_3 = -4i`. So

`f(x) = a(x - x_(1))*(x - x_(2))*(x-x_3)`

`f(x) = a(x - (-5))*(x - 4i)*(x - (-4i))`

`f(x) = a(x + 5)(x - 4i)(x + 4i)`

`f(x) = a(x + 5)(x^2 - 16i^2)`

Since
`i^2 = -1`

`f(x) = a(x + 5)(x^2 + 16)`

`f(x) = a(x^3 + 5x^2 + 16x + 80)`

Has a value of 40 when x=3

We use this to find a, which means that we have
`x = 3, f(x) = 40`. So

`a(3^3 + 5*3^2 + 16*3 + 80) = 40`

`200a = 40`

`a = (40)/(200) = (1)/(5) = 0.2`

So

`f(x) = 0.2(x^3 + 5x^2 + 16x + 80)`