Solution:
(x2+1)=(x−2)(x+2)
Let f(x)=x3+ax2+bx+c ; a,b,c∈{1,2,...,10}
f(x) is divisible by x2+1 iff f(i)=0,f(−i)=0
∴i3+ai2+bi+c=0
⇒−i−a+bi+c=0
⇒a=c,b=1
and −i3+ai2−bi+c=0
⇒i−a−bi+c=0
⇒a=c,b=1
∴ Total number of selection of triplets
(a,b,c)=(1,1,1), (2,1,2),...,(10,1,10)
Hence, number of such polynomials = 10.
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