Soal UN Matematika IPA 2015
11. Suku banyak P(x) bila dibagi (x2 - 5x + 6) sisanya yaitu (-2x + 3) dan bila suku banyak P(x) dibagi (x2 - x) sisanya (6x - 3), maka bila P(x-1) dibagi oleh (x2 - 4x + 3) memperlihatkan sisa ...
A. 3x - 4
B. x - 4
C. 6
D. -3x + 4
E. -x + 4
Jawaban : B
P(x) = (x2 - 5x + 6) . H(x) + S(x)
Pernyataan 1
P(x) = (x - 3)(x - 2) . H(x) + (-2x + 3)
bila x = 2, maka
P(2) = (2 - 3)(2 - 2) . H(2) + (-2 . 2 + 3)
P(2) = 0 + (-4 + 3)
P(2) = -1
bila x = 3, maka
P(3) = (3 - 3)(3 - 2) . H(3) + (-2 . 3 + 3)
P(3) = 0 + (-6 + 3)
P(3) = -3
Pernyataan 2
P(x) = (x2 - x) . H(x) + (6x - 3)
P(x) = x (x - 1). H(x) + (6x - 3)
bila x = 0, maka
P(0) = 0 (0 - 1). H(0) + (6 . 0 - 3)
P(0) = 0 + (0 - 3)
P(0) = -3
bila x = 1, maka
P(1) = 1 (1 - 1). H(1) + (6 . 1 - 3)
P(1) = 0 + (6 - 3)
P(1) = 3
---------------------------------------
misalkan, S(x) = ax + b
dan y = x-1
untuk y = 2, maka x = 3
untuk y = 0, maka x = 1
P(x-1) = (x2 - 4x + 3) . H(x) + S(x)
P(x-1) = ((x-1) - 3) . ((x-1) - 1) . H(x-1) + a(x-1) + b
P(3-1) = (3 - 3) (3 - 1) . H(3) + 3a + b
P(2) = 0 + 3a + b
- 1 = 3a + b ... (i)
P(x-1) = (x2 - 4x + 3) . H(x) + S(x)
P(x-1) = (x - 3) . (x - 1) . H(x) + ax + b
P(1-1) = (1 - 3) (1 - 1) . H(1) + a + b
P(0) = 0 + a + b
-3 = a + b ... (ii)
menggunakan eliminasi
- 1 = 3a + b
-3 = a + b
------------------
2 = 2a
a = 1
-3 = a + b
-3 = 1 + b
b = -3 - 1
b = -4
ax + b
= x - 4
11. Suku banyak P(x) bila dibagi (x2 - 5x + 6) sisanya yaitu (-2x + 3) dan bila suku banyak P(x) dibagi (x2 - x) sisanya (6x - 3), maka bila P(x-1) dibagi oleh (x2 - 4x + 3) memperlihatkan sisa ...
A. 3x - 4
B. x - 4
C. 6
D. -3x + 4
E. -x + 4
Jawaban : B
P(x) = (x2 - 5x + 6) . H(x) + S(x)
Pernyataan 1
P(x) = (x - 3)(x - 2) . H(x) + (-2x + 3)
bila x = 2, maka
P(2) = (2 - 3)(2 - 2) . H(2) + (-2 . 2 + 3)
P(2) = 0 + (-4 + 3)
P(2) = -1
bila x = 3, maka
P(3) = (3 - 3)(3 - 2) . H(3) + (-2 . 3 + 3)
P(3) = 0 + (-6 + 3)
P(3) = -3
Pernyataan 2
P(x) = (x2 - x) . H(x) + (6x - 3)
P(x) = x (x - 1). H(x) + (6x - 3)
bila x = 0, maka
P(0) = 0 (0 - 1). H(0) + (6 . 0 - 3)
P(0) = 0 + (0 - 3)
P(0) = -3
bila x = 1, maka
P(1) = 1 (1 - 1). H(1) + (6 . 1 - 3)
P(1) = 0 + (6 - 3)
P(1) = 3
---------------------------------------
misalkan, S(x) = ax + b
dan y = x-1
untuk y = 2, maka x = 3
untuk y = 0, maka x = 1
P(x-1) = (x2 - 4x + 3) . H(x) + S(x)
P(x-1) = ((x-1) - 3) . ((x-1) - 1) . H(x-1) + a(x-1) + b
P(3-1) = (3 - 3) (3 - 1) . H(3) + 3a + b
P(2) = 0 + 3a + b
- 1 = 3a + b ... (i)
P(x-1) = (x2 - 4x + 3) . H(x) + S(x)
P(x-1) = (x - 3) . (x - 1) . H(x) + ax + b
P(1-1) = (1 - 3) (1 - 1) . H(1) + a + b
P(0) = 0 + a + b
-3 = a + b ... (ii)
menggunakan eliminasi
- 1 = 3a + b
-3 = a + b
------------------
2 = 2a
a = 1
-3 = a + b
-3 = 1 + b
b = -3 - 1
b = -4
ax + b
= x - 4
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