Page 206 - aljaber

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(.)

:( ) ( )

10 = f (2) = a ×(2)2 + 2 + 1 = 4a + 3

.a = 7 . 4a = 7
:( )
4

( )4f (1) = 4 4 ×13 + 8 × 12 + 6 × 1 + 5 = 4 × 23 = 92 ()
( )5g(5) = 5 52 − 3 × 5 + 1 = 5 × 11 = 55 ()
()
. 4f (1) + 5g(5) = 92 + 55 = 147

:( )

(2x + 3)(x − 4) + (2x + 3)(x − 6) = (2x + 3)(x − 4 + x − 6)

= (2x + 3)(2x − 10)

.x1 + x2 = 5−3 = 7 .x2 = −3 x1 = 5
2 2 2
:( )
4 g(x) 3 f (x)

g(x )

( )g(x) = (x + b) x 3 + 3x 2 + 9x + 3

= x 4 + (3 + b)x 3 + (9 + 3b)x 2 + (3 + 9b)x + 3b

3+b = 4 ⇒b = 1

9 + 3b = 6p ⇒ 6p = 12 ⇒ p = 2

3 + 9b = 4q ⇒ 4q = 12 ⇒ q = 3

3b = r ⇒ r = 3

.(p + q)r = (2 + 3)× 3 = 15