多项式 f ( x ) {\displaystyle f(x)} 除以 ( x − 1 ) {\displaystyle (x-1)} ， ( x − 2 ) {\displaystyle (x-2)} 所得的余数分别为3和5，求 f ( x ) {\displaystyle f(x)} 除以 ( x − 1 ) ( x − 2 ) {\displaystyle (x-1)(x-2)} 所得的余式.
![\relative c'
<<
\new PianoStaff <<
\new Staff {\key bes \major r2 r4 <d f bes>8. <f bes d>16
<bes d f>4 r8 \tuplet 3/2 {<bes d f>16 <bes d f>16 <bes d f>16} <bes ees g>4 <bes d f>4~
<bes d f>4 \tuplet 3/2 { <bes d f>8 <bes d f>8 <bes d f>8 } \tuplet 3/2 { <bes ees g>8 <bes d f >8 <g bes ees>8 } <bes d f>8. <bes d f>16
<a c f>1~
<a c f>8 r8 <e g c>8. <g c e>16 <c e g>4~ <c e g>8 \tuplet 3/2 { <c e g>16 <c e g>16 <c e g>16}
<c f a>4 <c e g>2~ <c e g>16[ r32 <c e g>8 <c e g>8]
\tuplet 3/2 { <c f a>8 <c e g>8 <a d f>8 } <c ees g>8. <c ees g>16 <b d g>2~
<b d g>16[ r32 <c g'>8 <c g'>8] \tuplet 3/2 { <c a'>8[ <c g'>8 <a f'>8]}
}
\relative c
\new Staff {\clef "bass" \key bes \major r1 r2 r4 r8 <f, f'>8 <bes bes'>2~ <bes bes'>8 r8 r4 a'1 r1
r4 r8 g,8 c2~ c8 r8 r4 <g g'>2
g4 r4 r2.
}
>> % end PianoStaff
\relative c'
\new PianoStaff <<
\new Staff { \key bes \major
r2 r4 <d f bes>8. <f bes d>16
<bes d f>4 r8 \tuplet 3/2 {<bes d f>16 <bes d f>16 <bes d f>16} <bes ees g>4 <bes d f>4~
<bes d f>4 \tuplet 3/2 { <bes d f>8 <bes d f>8 <bes d f>8 } \tuplet 3/2 { <bes ees g>8 <bes d f >8 <g bes ees>8 } <bes d f>8. <bes d f>16
<a c f>1~
<a c f>8 r8 <e g c>8. <g c e>16 <c e g>4~ <c e g>8 \tuplet 3/2 { <c e g>16 <c e g>16 <c e g>16}
<c f a>4 <c e g>2~ <c e g>16[ r32 <c e g>8 <c e g>8]
\tuplet 3/2 { <c f a>8 <c e g>8 <a d f>8 } <c ees g>8. <c ees g>16 <b d g>2
}
\relative c
\new Staff { \key bes \major \clef "bass"
r1
r2 r4 r8 f,,8
bes2~ bes8 r8 r4
f''1
r1
r4 r8 g,8 <c c,>2~ <c c,>8 r8 r4 <g f' g>2~
\time 5/4
<g f' g>8 r8 r4 r2 r4
}
>>
>>](http://upload.wikimedia.org/score/e/h/ehokv9syhlelqpn43a232501j7vsujs/ehokv9sy.png)
