jika suku banyak f(x) 3x3+10x2+(a-1)x+9 dibagi (3x+1) sisanya adalah 7. tentukan nilai a

TEOREMA SISA

f(x) = 3x³ + 10x² + (a - 1)x + 9

3x + 1 = 0
3x = 0 - 1
3x = -1
x = -1/3

f(-1/3) = 7
3(-1/3)³ + 10(-1/3)² + (a - 1)(-1/3) + 9 = 7
3(-1/27) + 10(1/9) - 1/3 a + 1/3 + 9 = 7
-1/9 + 10/9 - 1/3 a + 28/3 = 7
9/9 - 1/3 a + 28/3 = 7
1 -1/3 a + 28/3 = 7
-1/3 a + 31/3 = 7
-1/3 a = 7 - 31/3
-1/3 a = 21/3 - 31/3
-1/3 a = -10/3
a = -10/3 (-3) = 10

maka nilai a = 10