Difference between revisions of "Conjugate Root Theorem"
AlcumusGuy (talk | contribs) |
Claudeaops (talk | contribs) (Additional Theorem) |
||
Line 1: | Line 1: | ||
=Theorem= | =Theorem= | ||
The Conjugate Root Theorem states that if <math>P(x)</math> is a polynomial with real coefficients, and <math>a+bi</math> is a root of the equation <math>P(x) = 0</math>, where <math>i = \sqrt{-1}</math>, then <math>a-bi</math> is also a root. | The Conjugate Root Theorem states that if <math>P(x)</math> is a polynomial with real coefficients, and <math>a+bi</math> is a root of the equation <math>P(x) = 0</math>, where <math>i = \sqrt{-1}</math>, then <math>a-bi</math> is also a root. | ||
+ | A similar theorem states that if <math>P(x)</math> is a polynomial with rational coefficients and <math>a+b\sqrt{c}</math> is a root of the polynomial, then <math>a-b\sqrt{c}</math> is also a root. | ||
==Uses== | ==Uses== |
Revision as of 14:27, 1 January 2014
Theorem
The Conjugate Root Theorem states that if is a polynomial with real coefficients, and is a root of the equation , where , then is also a root. A similar theorem states that if is a polynomial with rational coefficients and is a root of the polynomial, then is also a root.
Uses
This has many uses. If you get a fourth degree polynomial, and you are given that a number in the form of is a root, then you know that in the root. Using the Factor Theorem, you know that is also a root. Thus, you can multiply that out, and divide it by the original polynomial, to get a depressed quadratic equation. Of course, it doesn't have to be a fourth degree polynomial. It could just simplify it a bit.
This article is a stub. Help us out by expanding it.