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

**Solution:**

The conjugate root theorem states that if the complex number a + bi is a root of a polynomial

P(x) in one variable with real coefficients, then the complex conjugate a - bi is also a root of

that polynomial.

Given, the roots are i, -2 and 2.

The polynomial is the product of its factors,

So, f(x) = (x - i) (x + i) (x - 2) (x + 2)

f(x) = (x^{2} - i^{2})(x^{2} - 2^{2} )

Using the algebraic identity (a^{2} - b^{2}) = (a + b) (a - b)

We know that i^{2} = -1

f(x) = (x^{2} + 1)(x^{2 }- 4)

f(x) = x^{4} - 4x^{2} + x^{2} - 4

f(x) = x^{4} - 3x^{2} - 4

Therefore, the polynomial function of lowest degree is f(x) = x^{4} - 3x^{2} - 4.

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

**Summary:**

The polynomial function of lowest degree with lead coefficient 1 and roots i, -2, and 2 is f(x) =x^{4} - 3x^{2} - 4.

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