Question

What is the polynomial function of lowest degree with lead coefficient 1 and roots i, –2, and 2?

A. f(x)=x^3-x^2-4x+4
B. f(x)=x^4-3x^2-4
C. f(x)=x^4+3x^2-4
D. f(x)=x^3+x^2-4x-4

Answer 1

The answer is B f(x)=x^4-3x^2-4

Answer 2

`\boxed{\boxed{B.\ f(x)=x^4-3x^2-4}}`

Explanation:

Complex Conjugate Root Theorem-

If
`a + bi` is a root of a polynomial P with a and b real numbers, then its complex conjugate
`a-bi` is also a root of P.

So, all roots of the polynomial function are
`i,-i, 2,-2`

Hence, the function will be,

`f(x)=(x-i)(x+i)(x-2)(x+2)`

`=[(x-i)(x+i)]\cdot[(x-2)(x+2)]`

`=(x^2-i^2)(x^2-2^2)`

`=(x^2+1)(x^2-4)`

`=x^2x^2+x^2\left(-4\right)+1\cdot \:x^2+1\cdot \left(-4\right)`

`=x^2x^2-4x^2+1\cdot \:x^2-1\cdot \:4`

`=x^4-3x^2-4`

The leading coefficient in this case is 1, so the function is,

`f(x)=x^4-3x^2-4`