If 2x3+4x2+2ax+b is exactly divisible by x2−1 Then the value of a and b respectively will be
1,2
−1,−4
1,−2
−1,4
A
−1,4
B
−1,−4
C
1,2
D
1,−2
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Solution
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Since f(x)=2x3+4x2+2ax+b is exactly divisible by x2−1=(x−1)(x+1) ∴f(1)=0 and f(−1)=0 These give 2+4+2a+b=0 or 2a+b+6=0 .....(i) and −2+4−2a+b=0 or 2a−b−2=0 ....(ii) Solving equations (i) and (ii) we get a=−1,b=−4
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