Page 187 - aljaber

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2 . f (−2) = 0 f (2) = 0 f (x) = x2 − 4
. −2

f (x ) = ax 4 − bx 2 + x + 5 [AHSME 1995]( )

. f (3) f (−3) = 2

f (−3) = 2 ⇒ a(−3)4 − b(−3)2 − 3 + 5 = 2 ⇒ 81a − 9b = 0

. f (3) = a(3)4 − b(3)2 + 3 + 5 = (81a − 9b) + 8 = 0 + 8 = 8

[Operations on Poynomials] (.)

g(x ) f (x)
f (x ) = 2x 2 − 36x + 9
.

g(x) = 2x 3 + 4x 2 + 5x − 1

f (x) + g(x) = 2x 3 + 6x2 − 31x + 8
f (x) − g(x) = −2x 3 − 2x 2 − 41x + 10

. ()

g(x) = x 2 + 2x + 4 f (x) = x − 2
f (x)g(x) = (x − 2)(x 2 + 2x + 4)

= x(x 2 + 2x + 4) − 2(x 2 + 2x + 4)