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| − | =Theorem=
| + | #REDIRECT[[Complex Conjugate Root 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.
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| − | 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.
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| − | ==Proof==
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| − | Suppose that <math>P(a + bi) = 0</math>. Then <math>\overline{P(a + bi)} = 0</math>. However, we know that <math>\overline{P(a + bi)} = \overline{P}(\overline{a + bi}) = P(a - bi)</math>, where we define <math>\overline{P}</math> to be the polynomial with the coefficients replaced with their [[complex conjugate|complex conjugates]]; we know that <math>\overline{P} = P</math> by the assumption that <math>P</math> has real coefficients. Thusly, we show that <math>P(a - bi) = 0</math>, and we are done.
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| − | ==Uses==
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| − | This has many uses. If you get a fourth degree polynomial, and you are given that a number in the form of <math>a+bi</math> is a root, then you know that <math>a-bi</math> in the root. Using the [[Factor Theorem]], you know that <math>(x-(a+bi))(x-(a-bi))</math> is also a root. Thus, you can multiply that out, and divide it by the original polynomial, to get a [[quadratic formula|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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| − | {{stub}}
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| − | [[Category:Theorems]]
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| − | [[Category:Algebra]]
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| − | [[Category:Polynomials]]
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