# Math Help - Polynomials

1. ## Polynomials

What is the degree of the simplest polynomial with the zeroes:√2, 2√2 + i, and 2 + 2i ??

2. My guess would be 6 since the simplest polynomial that I can come up with that has all of these as roots is (x² - 2)(x² - 4√2 x + 9)(x² - 4x + 8).

3. This means that $x - \sqrt{2}$, $x - (2 \sqrt{2} + \pi)$, $x - (2 + 2 \pi)$ are all factors of this polynomial. Therefore the simplest polynomial that has all these roots is :

$(x - \sqrt{2})(x - (2 \sqrt{2} + \pi))(x - (2 + 2 \pi)) = 0$

Or, equivalently :

$(x - \sqrt{2})(x - 2 \sqrt{2} - \pi)(x - 2 - 2 \pi) = 0$

And the degree of this polynomial is, trivially, 3 (can be checked by expanding).

4. Originally Posted by VMM
What is the degree of the simplest polynomial with the zeroes:√2, 2√2 + i, and 2 + 2i ??
Roots or zeros of polynomials of degree greater than 2 - Topics in precalculus