Jawaban:

f(x) = hasil bagi × pembagi + sisa

tujuan teorema sisa :

1. cari nilai x yg membuat hasil bagi × pembagi = 0 (akar pembagi)
2. shg f(x) = 0 + sisa
3. f(x) = sisa

mis no.1 ==> pembagi = (x-2)

agar f(x) = sisa ==> hasil bagi x pembagi = 0

H(x) × (x-2) = 0

substitusi x dgn akar (x-2) ==> H(2) × (2-2) =0

shg ==> f(2) = sisa

2⁴-3(2)³+5(2)²-4(2)+1 = sisa

5 = sisa

2. akar pembagi = -3

f(-3) = H(3) × (-3+3) + sisa

f(-3) = 0 + sisa

2(-3)³-4(-3)²+8(-3)-5 = sisa

-119 = sisa

3. akar pembagi = ½

f(½) = sisa

8(½)⁴-2(½)³+5(½)-6 = sisa

sisa = -3¼

4.

pembagi (x+3)(x-1) ==> akar -3 & 1

sisa : 1 tingkat dibawah pembagi ==> ax+b

f(-3) = sisa ax+b

6(-3)⁴-3(-3)²+3(-3)+6 = a(-3)+b

-3a+b = 456 … [1]

f(1) = sisa ax+b

6(1)⁴-3(1)²+3(1)+6 = a(1)+b

a+b = 12 … [2]

[1] & [2]

-3a+b = 456

a+b = 12

------------------ (-)

-4a = 444

a = -111

b = 123

sisa = -111x + 123