餘式定理與因式定理

出自高材生

跳轉到: 導航, 搜索

3 餘式定理與因式定理

  • 觀念說明
    • 餘式定理
      • 已知：\left\{ \begin{align} & f\left( x \right)\text{ is a polynomial} \\ & f\left( x \right)=\left( ax-b \right)\times q\left( x \right)+r \\ \end{align} \right. ，求證：r=f\left( \frac{b}{a} \right)
      • 證明：f\left( x \right)=\left( ax-b \right)\times q\left( x \right)+r\Rightarrow f\left( \frac{b}{a} \right)=\left( a\times \frac{b}{a}-b \right)\times q\left( \frac{b}{a} \right)+r=0\times q\left( \frac{b}{a} \right)+r=r\#
    • 因式定理
      • 已知：\left\{ \begin{align} & f\left( x \right)\text{ is a polynomial} \\ & \left( ax-b \right)|f\left( x \right) \\ \end{align} \right. ，求證：f\left( \frac{b}{a} \right)=0
      • 證明：\left( ax-b \right)|f\left( x \right)\Rightarrow f\left( x \right)=\left( ax-b \right)q\left( x \right)\Rightarrow f\left( \frac{b}{a} \right)=\left( a\times \frac{b}{a}-b \right)\times q\left( \frac{b}{a} \right)=0\times q\left( \frac{b}{a} \right)=0\#


範例6-9　|　例1-11

取自"http://www.come2pass.com/wiki/index.php/%E9%A4%98%E5%BC%8F%E5%AE%9A%E7%90%86%E8%88%87%E5%9B%A0%E5%BC%8F%E5%AE%9A%E7%90%86"