Question

Suku banyak R(x) dibagi (2X - 1) dan (3X + 2) berturut-turut bersisa 2 dan -3.
Suku banyak N(x) dibagi (2X - 1) dan (3X + 2) berturut-turut bersisa -2 dan 6.
Tentukan pembagian suku banyak H(x) = R(x).N(x) oleh (2X -1).(3X +2)!

Answer 1

R(2x-1) = 2 --> r(1/2) = 2
r(3x+2) = -3 -->r(-2/3) = -3

n(2x-1) = -2 --> n(1/2) = -2
n(3x+2) = 6 --> n(-2/3) = 6

h(x) = r(x). n(x)
h(x) ; (2x-1)(3x+2) --> sisa s(x) = ax+ b

s(1/2) = h(1/2) = r(1/2). n(1/2) = - 4
1/2 a + b = - 4 ---> a + 2b = - 8 .....(1)

s(-2/3) = h(-2/3) = r(-2/3) n(-2/3) = -18
-2/3 a + b = - 18 --> -2a + 3b = - 54...(2)

eliminasi (1) dan (2)
a + 2b = -8
-2a + 3b = - 54
..
2a + 4b  =  -16
-2a + 3b = -54
....7b = - 70
b = -10 
dan  a +  2b = - 8
a - 20 = - 8
a = 12

s(x)= ax + b  ----> s(x) = 12x - 10