if zeroes of a quadratic polynomial x^2+(a+1)x+b are 2 and -3 find the value of a and b
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Answer:a=0 and b= -6
Step-by-step explanation:
Put x=2
(2 x 2) + (a+1)2 + b
=6+2a+b ........(1)
Put x= -3
(-3 x -3) + (a+1)-3 +b
=6-3a+6.........(2)
On putting both equations we get
6+2a+b = 6-3a+b
2a+3a=0
5a=0
a=0
On substituting the value of a, we get
6 x 2(0) + b=0
b=-6
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