Suku banyak berderajat 3 jika dibagi (x² + 2x - 3) bersisa (3x - 4) dan jika dibagi (x² - x - 2) bersisa (2x + 3) suku banyak tersebut adalah

X² + 2x - 3 = 0

(x + 3) (x - 1) = 0

x = -3,x = 1

f(-3) = 3(-3) - 4

= -9 - 4 = -13

f(1) = 3 . 1 - 4

= 3 - 4 = -1

f(x) = P(x) . H(x) + s(x)

f(-3) = ((-3)² - (-3) - 2) (ax + b) + 2x + 3

-13 = (9 + 3 - 2) (-3a + b) + 2 (-3) + 3

-13 = 10(-3a + b) - 3

-13 = -30a + 10b - 3

-30a + 10b = -13 - (-3)

-30a + 10b = -13 + 3

-30a + 10b = -10 ...(1)

f(1) = (1² - 1 - 2) (ax + b) + 2x + 3

-1 = -2(a + b) + 2 . 1 + 3

-1 = -2a - 2b + 2 + 3

-1 = -2a - 2b + 5

-2a - 2b = -1 - 5

-2a - 2b = -6 ...(2)

-30a + 10b = -10

-2a - 2b = -6

-30a - 30b = -90

______________-

40b = 80

b = 80/40 = 2

-2a - 2b = -6

-2a - 2(2) = -6

-2a - 4 = -6

-2a = -6 - (-4)

-2a = -6 + 4

-2a = -2

a = -2/(-2) = 1

f(x) = (x² - x - 2) (x + 2) + 2x + 3

f(x) = x³ + 2x² - x² - 2x - 2x - 4 + 2x + 3

f(x) = x³ + x² - 2x - 1

maka suku banyak tersebut adalah

x³ + x² - 2x - 1✓