16a . E(x) = x³ -2x² -x + 2 = x² ( x -2) - ( x -2)
= ( x -2) · ( x² -1 )
= ( x -2) · ( x -1) · ( x +1)
b. 5( x² - 1) / ( x -2) · ( x² -1) = 5 / ( x -2) ∈ Z
daca x - 2 = - 5 x = - 3
x - 2 = - 1 x = 1
x - 2 = 1 x = 3
x - 2 = 5 x = 7
x ∈ { - 3 ; 1 ; 3 ; 7 }
17a. E(x) = x²( x + 1) - 4(x + 1) = ( x + 1)·( x² -4)
= ( x + 1) · ( x -2) ·( x +2)
F(x) = x²·( x + 2) - ( x + 2) = ( x + 2) ·( x² - 1)
= ( x + 2) · ( x -1) ·( x +1)
b. E(x) / F(x) = ( x -2 ) / ( x -1)
c. E(x) + F(x) = 0
( x + 1) ·( x + 2) · [ x - 2 + x -1] = 0
( x + 1) · ( x +2) · ( 2x - 3 ) = 0
daca x=-1 x = -2 x = 3 /2
A = { - 2 ; - 1 ; 3 / 2 }
18 . E(x) = ( x² + x)² + 2·( x² + x) - 4·( x² + x) - 8 + 8
= ( x² + x)² - 2 · ( x² + x) = ( x² + x)·( x² + x - 2)
= x · ( x + 1) · ( x - 1) · ( x +2)
F(x) = ( x² -x )² - 2·(x² -x) - 4 ·( x² -x) + 8 - 8
= ( x² - x)² - 6·( x² - x) =( x² -x) ·( x² - x - 6)
= x ·( x -1) ·( x -3) ·( x +2)
b. E(x) / F(x) = ( x +1) / ( x -3)
c. ( x + 1) / ( x -3) ∈ Z daca exista un divizor d :
d | x + 1
d | x - 3
--------------------------------
atunci d | x+ 1 - x + 3
divizorul este d = 4
x - 3 = - 4 x = -1
x - 3 = -2 x =1
x - 3 = -1 x =2
x -3 = 1 x =4
x -3 = 2 x = 5
x - 3 = 4 x = 7
