已知\(f(x)\)是三次實係數多項式，且\(f(11)=2012\)，\(f(21)=-2013\)，\(f(31)=2014\)，設\( \displaystyle g(x)=\frac{(x-21)(x-31)}{(11-21)(11-31)}\cdot 2012+\frac{(x-11)(x-31)}{(21-11)(21-31)}\cdot (-2013)+\frac{(x-11)(x-21)}{(31-11)(31-21)}\cdot 2014\)且\(r(x)\)為\( f(x) \)除以\( (x-11)(x-21)(x-31) \)之餘式，請問下列哪些選項是正確的？
(A)\(r(11)=2012\)
(B)\(g(41)=-2015\)
(C)方程式\(f(x)=0\)恰有3個相異實根
(D)方程式\( f(x)-g(x)=0 \)恰有3個相異實根
(E)方程式\( r(x)=0 \)恰有2個相異實根

可以請問一下，這題的答案是acde
但我不會的選項是bcde
有高手可以教一下為什麼？
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\(\begin{align}
  & f\left( x \right)=\left( x-11 \right)\left( x-21 \right)\left( x-31 \right)q\left( x \right)+r\left( x \right) \\
& r\left( 11 \right)=f\left( 11 \right)=2012 \\
& r\left( 21 \right)=f\left( 21 \right)=-2013 \\
& r\left( 31 \right)=f\left( 31 \right)=2014 \\
& r\left( x \right)=g\left( x \right)=\frac{\left( x-21 \right)\left( x-31 \right)}{\left( 11-21 \right)\left( 11-31 \right)}\times 2012+\frac{\left( x-11 \right)\left( x-31 \right)}{\left( 21-11 \right)\left( 21-31 \right)}\times \left( -2013 \right)+\frac{\left( x-11 \right)\left( x-21 \right)}{\left( 31-11 \right)\left( 31-21 \right)}\times 2014 \\
&  \\
& g\left( 41 \right)=2012+6039+6042=14093 \\
\end{align}\)


大略描出\(\left( 11,2012 \right),\left( 21,-2013 \right),\left( 31,2014 \right)\)這三個點，就知道\(f\left( x \right)=0\)至少有二實根，由於它是三次，故有三實根

\(f\left( x \right)-g\left( x \right)=0\)有11、21、31這三個實根

大略描出\(\left( 11,2012 \right),\left( 21,-2013 \right),\left( 31,2014 \right)\)這三個點，就知道\(r\left( x \right)=0\)有二實根

[ 本帖最後由 thepiano 於 2015-7-19 07:31 AM 編輯 ]

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