Difference between revisions of "Conjugate Root Theorem"
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− | 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 [[coeffi ient|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. | 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. | ||
Revision as of 20:43, 27 February 2020
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.
Proof
Suppose that . Then . However, we know that , where we define to be the polynomial with the coefficients replaced with their complex conjugates; we know that by the assumption that has real coefficients. Thusly, we show that , and we are done.
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.
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