Polynomials

• May 15th 2010, 05:00 PM
VMM
Polynomials
What is the degree of the simplest polynomial with the zeroes:√2, 2√2 + i, and 2 + 2i ??
• May 15th 2010, 07:24 PM
NowIsForever
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).
• May 15th 2010, 07:29 PM
Bacterius
This means that $\displaystyle x - \sqrt{2}$, $\displaystyle x - (2 \sqrt{2} + \pi)$, $\displaystyle x - (2 + 2 \pi)$ are all factors of this polynomial. Therefore the simplest polynomial that has all these roots is :

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

Or, equivalently :

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

And the degree of this polynomial is, trivially, 3 (can be checked by expanding).
• May 16th 2010, 06:52 AM
skeeter
Quote:

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
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