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P(x)=ax²+bx+c
x²-x-6 => (x-3)(x+2) = (5x-2)
P(3)=9a+3b+c=13 ...(1)
P(-2)=4a-2b+c=-12 ...(2)
x²-2x-3 => (x-3)(x+1) = (3x+4)
P(-1)=a-b+c=1 ...(3)
P(3)=9a+3b+c=13 ...(4)
• Eliminasi pers (1) dan (2)
9a+3b+c = 13
4a-2b+c = -12
__________ _
5a+5b = 25
a+b = 5 ...(5)
• Eliminasi pers (3) dan (4)
a-b+c = 1
9a+3b+c = 13
__________ _
-8a-4b = -12
2a+b = 3 ...(6)
• Eliminasi pers (5) dan (6)
a+b = 5
2a + b = 3
_______ _
-a = 2
a = -2
• substitusikan a=-2 ke pers (5)
a+b = 5
-2 + b = 5
b = 7
• substitusikan a = -2 dan b = 7 ke pers (3)
a-b+c = 1
-2-7+c = 1
c = 10
maka, suku banyak tersebut adlah :
-2x²+7x+10