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{{ printedBook.courseTrack.name }} {{ printedBook.name }} Let $p(x)$ be a polynomial function whose coefficients are real. If a complex number $a+bi$ is a root of $p(x),$ then the root's complex conjugate, $a−bi,$ is also a root of $p(x).$

$p(a+bi)=0⟹p(a−bi)=0 $

Let $p(x)$ be a polynomial with real coefficients $c_{k}.$ $p(x)=k=0∑n c_{k}x_{k} $ Using the properties of conjugation, the conjugate of the polynomial evaluated at $a+bi$ can be rewritten.

$p(a+bi) =k=0∑n c_{k}(a+bi)_{k} $ | Conjugate of the polynomial evaluated at $a+bi$ |

$p(a+bi) =k=0∑n c_{k}(a+bi)_{k} $ | Conjugate of a sum is the sum of the conjugates |

$p(a+bi) =k=0∑n c_{k} (a+bi)_{k} $ | Conjugate of a product is the product of the conjugates |

$p(a+bi) =k=0∑n c_{k} (a+bi )_{k}$ | Conjugate of a power is the power of the conjugate |

$p(a+bi) =k=0∑n c_{k}(a+bi )_{k}$ | Conjugate of a real number is itself |

$p(a+bi) =k=0∑n c_{k}(a−bi)_{k}$ | The conjugate of $a+bi$ is $a−bi$ |

$p(a+bi) =p(a−bi)$ | The polynomial evaluated at $a−bi$ |

It is given that $a+bi$ is a root of the polynomial $p(x).$ $p(a+bi)=0 $ The conjugate of the real number $0$ is $0,$ so the conjugate of the polynomial evaluated at $a+bi$ is $0.$ $p(a+bi) =0 $ Comparing this with the equality in the last line of the table shows that $a−bi,$ the conjugate of $a+bi$ is also a root of the polynomial. $p(a+bi) p(a+bi) =p(a−bi)=0 ⟹p(a−bi)=0 $