if p(x)=x^3-ax^2+bx+3 leaves a remainder-19 when divided by (x+2)and a remainder 17 when divided by(x-2)prove that a+b=6

Asked by Anishma Joseph | 15th Jul, 2013, 06:56: PM

Expert Answer:

p(x)=x^3-ax^2+bx+3
 
p(-2) = -19
-8 -4a -2b + 3 = -19
-4a -2b = -14 (2)
 
Also, p(2) = 17
8 -4a + 2b + 3 = 17
11 -4a + 2b = 17
-4a + 2b = 6 (2)
 
Subtracting1 and 2
2b = 20
b = 5
 
hence -4a = 6 - 2*5 = -4
a = 1
 
Hence a + b = 1+5= 6
hence proved. 
 

Answered by  | 15th Jul, 2013, 10:43: PM

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