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( x  ax  bx  c )  ( x  1)( x  px  q )  D

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[ 東京工業大学 1978 年 １ ]

定数

a b c p q D , , , , ,

に対して，次の等式

3 2 2 2 2 2

( xaxbxc )  ( x  1)( xpxq )  D

がすべてのxについて成り立つとき，

D

の値を求めよ。

3 2 2

( xaxbx c  )

6 5 2 4 3 2 2 2

2 ( 2 ) (2 2 ) (2 ) 2

x ax a b x ab c x ac b x bcx c

         

( x

2

 1)( x

2

pxq )

2

D

6 5 2 4 3 2 2 2 2

2 ( 2 1) (2 2 ) ( 2 ) 2

x px p q x pq p x q q p x pqx q D

            

これらがすべてのxについて等しくなることから，係数を比較して

2 a  2 p

…①

a

2

 2 bp

2

 2 q  1

…②

2 ab  2 c  2 pq  2 p

…③

2 ac b

2

q

2

 2 qp

2 …④

2 bc   2 pq

…⑤

c

2

   q

2

D

…⑥

①，②より

1

, 2

ap b   q

…⑦

⑦を③に代入して

1

c   2 p

…⑧ ④，⑦，⑧より

1 3

4 , 4

q   b  

…⑨

⑤，⑧，⑨より

p   c 0

…⑩ ⑥，⑨，⑩より

1

D  16

参照

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