ada yang bisa bantu? pakai cara ya
Matematika
shofielathifah
Pertanyaan
ada yang bisa bantu? pakai cara ya
1 Jawaban
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1. Jawaban whongaliem
teorema sisa : f (x) : (x - a) ⇒ sisa f (a)
pembagi : x - 2 ⇒ f (2) = sisa
f (2) = 5
pembagi : x² - x - 6 = (x + 2) (x - 3)
f x) = p (x) .h (x) + sisa
= (x + 2) (x - 3) . h(x) + 3x - 1
f( - 2) = (- 2 + 2) ( - 2 - 3) .h ( - 2) + 3 ( - 2) - 1
= 0 ( - 5) . h (x) - 6 - 1
= 0 - 6 - 1
= - 7
pembagi : x² - 4 = (x - 2)(x + 2)
f (x) = p (x) . h (x) + sisa
= (x - 2) (x + 2) h (x)+ ax + b
f (2) = (2 - 2) (2 + 2) . h (x) + a (2) + b
5 = (0 ) (4) .h (2) + 2a + b
5 = 0 + 2a + b
5 = 2a + b ....... pers I
f ( - 2) = ( - 2 - 2) ( - 2 + 2) .h ( - 2) + a ( - 2) + b
- 7 = ( - 4) (0) h ( - 2) - 2a + b
- 7 = 0 - 2a + b
- 7 = - 2a + b .... pers II
elemenasi pers I dan II
5 = 2a + b
- 7 = - 2a + b
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12 = 4a
a = 12/4
a = 3
5 = 2a + b
5 = 2 (3) + b
5 = 6 + b
b = 5 - 6
b = - 1
jadi , sisa = ax + b
= 2x - 1