Page 195 - aljaber
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. sr x 2 + 2x + 5 = 0
x 2 − 2x + 5 = 0 . −s −r
. x 2 − 2x + 5
( )( )x2 + 2x + 5 x2 − 2x + 5 = x 4 + px 2 + q
x 4 + 6x 2 + 25 = x 4 + px 2 + q
.q = 25 p = 6
f (−5) f (5) = 3 f (x ) = ax 3 + bx − 7 ()
.
f (5) = 53a + 5b − 7 ⇒ 53a + 5b = 10
( ). f (−5) = −53a − 5b − 7 = − 53a + 5b − 7 = −10 − 7 = −17
d≠0 c≠0 [MAΘ 1991]( )
α1 + α2 = 0 α3 α2 α1 4x 3 − 12x 2 + cx + d = 0
.d
c
4x 3 − 3x 2 + c x + d = 4(x − α1 )(x − α2 )(x − α3 )
4 4
( )= 4 x 2 − (α1 + α2 )x + α1α2 (x − α3 )
( )= 4 x 2 + α1α2 (x − α3 )
( )= 4 x 3 − α3x 2 + α1α2x − α1α2α3