PEMBAHASAN
POLinomial
soal pertama
f(x) = 2x³ + (a + 3)x² - 18x - 3
f(4) = 5
2.4³ + (a + 3).4² - 18.4 - 3 = 5
128 + 16a + 32 - 72 - 3 - 5 = 0
16a + 80 = 0
16a = -80
a = -5
Nilai a = -5
•
soal kedua
diketahui fungsi f(x)
f(x) = (x² - 2x - 15) H(x) + (3x + 2)
f(x) = (x - 5)(x + 3) H(x) + (3x + 2)
f(5) = 3.5 + 2 = 17
f(-3) = 5(-3) - 4 = -19
f(x) = (x² - 2x - 3) H(x) + (5x - 4)
f(x) = (x - 3)(x + 1) H(x) + (5x - 4)
f(3) = 5.3 - 4 = 11
f(-1) = 5(-1) - 4 = -9
f(x) = (x² - 8x + 15) H(x) + (ax + b)
f(x) = (x - 5)(x - 3) H(x) + (ax + b)
f(5) = 17
5a + b = 17 ... (1)
f(3) = 11
3a + b = 11 ... (2)
Eliminasi substitusi persamaan (1) dan (2) :
a = 3 dan b = 2
Maka, sisa pembagian f(x) oleh (x² - 8x + 15) :
ax + b = 3x + 2
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
Verified answer
PEMBAHASAN
POLinomial
soal pertama
f(x) = 2x³ + (a + 3)x² - 18x - 3
f(4) = 5
2.4³ + (a + 3).4² - 18.4 - 3 = 5
128 + 16a + 32 - 72 - 3 - 5 = 0
16a + 80 = 0
16a = -80
a = -5
Nilai a = -5
•
soal kedua
diketahui fungsi f(x)
•
f(x) = (x² - 2x - 15) H(x) + (3x + 2)
f(x) = (x - 5)(x + 3) H(x) + (3x + 2)
f(5) = 3.5 + 2 = 17
f(-3) = 5(-3) - 4 = -19
•
f(x) = (x² - 2x - 3) H(x) + (5x - 4)
f(x) = (x - 3)(x + 1) H(x) + (5x - 4)
f(3) = 5.3 - 4 = 11
f(-1) = 5(-1) - 4 = -9
•
f(x) = (x² - 8x + 15) H(x) + (ax + b)
f(x) = (x - 5)(x - 3) H(x) + (ax + b)
f(5) = 17
5a + b = 17 ... (1)
f(3) = 11
3a + b = 11 ... (2)
Eliminasi substitusi persamaan (1) dan (2) :
a = 3 dan b = 2
Maka, sisa pembagian f(x) oleh (x² - 8x + 15) :
ax + b = 3x + 2