(2x³+ax²+11x+a+3) is exactly divisible by (2x-1)
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2x-1=0
x=1/2
Now, substituting the value of x in 2x³+ax²+11x+a+3, we get-
2(1/2)³+a(1/2)²+11(1/2)+a+3=0
1/4+a/4+11/2+a+3=0
35/4+5a/4=0
thus, a= -35*4/5*4= -7
x=1/2
Now, substituting the value of x in 2x³+ax²+11x+a+3, we get-
2(1/2)³+a(1/2)²+11(1/2)+a+3=0
1/4+a/4+11/2+a+3=0
35/4+5a/4=0
thus, a= -35*4/5*4= -7
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