Tentukan sisa pembagin polinomial berikut dengan menerapkan teorema sisa a.(x⁴-16):(x²-2) b.(x⁴-3x²+7) : (2x-1) c.(x^7+3x^5+1) : (x²-x) 2.jika polinomial f(x)=

Suku Banyak / Teorema Sisa

1. Sisa pembagian
a) (x⁴  -16) : (x² -2) = 
f(x) = x⁴ - 16
p(x) = x²  - 2
sisa = s(x)= ax + b
x² - 2 = 0  --> x² = 2
x= √2 atau x = - √2
s(x) = f(x)
s(√2) = f(√2) --> √2  (a)  + b = (√2)⁴ - 16
a√2 + b=  -12.....(1)

s(-√2)= f(-√2) --> -√2 (a) + b = (-√2)⁴ -16
-a√2 + b = -12 ...(2)

eliminasi (1) dengan (2)
a√2 + b = -12
-a√2 + b = -12
jumlahkan
2b = -24
b  = - 12
.
a√2 + b = -12 --> a√2 - 12 = - 12
a√2 = 0
a = 0
sisa = s(x) = ax + b --> s(x)= 0(x)  - 12
s(x) = -12

b.
f(x) = x⁴ - 3x² + 7
p(x) = 2x -1
s(x)= c
2x -1= 0
x = 1/2
s(x) = f(x)
s(1/2) = f(1/2)
s(1/2) = (1/2)⁴ -3(1/2)² + 7
s(1/2) =101/16 
sisa = 6 ⁵/₁₆

c.
f(x) = x⁷ + 3x⁵ + 1
p(x) =x² - x
s(x) = ax + b
x² - x = 0 --> x(x -1)= 0
x = 0  ataux = 1
s(x) = f(x)
s(0) = f(0) --> a(0) + b = 0  + 0 + 1
b = 1
s(1) = f(1) --> a(1) + b  = 1 + 3 + 1
a + b = 5 --> a + 1 = 5 
a = 4
s(x) = ax + b 
s(x) = 4x + 1

2)
f(x) = x⁴ + 2x³ + kx² - 4x - 3
g(x) = 3x⁵ + 5x² - kx - 6
p(x) = x - 1
x - 1= 0
x = 1
sisa yang sama maka  f(1) = g(1)
1 + 2 + k - 4 - 3 = 3 + 5 - k - 6
k - 4 = -k + 2
2k = 6
k = 3

3. 
f(x)= x⁴ + x² - 20
habis dibagi x - 2 --> f(2) = 0
2⁴ + 2² - 20  = 16 + 4 - 20= 0
terbukti
habis dibagi x + 2--> f(-2) = 0
(-2)⁴  + (-2)² - 20 = 16 + 4 - 20 = 0
terbukti

(x⁴ + x² - 20) = (x-2)(x+2)(x²  + 5) 
hasil bagi (x⁴ + x² -20) : (x-2) = (x+2)(x² +5)
hasil bagi (x⁴ + x² -20) : (x+2) = (x -2)(x² + 5)