Let 2x5+5x4+10x3+10x2+10x+10=f(x)
Now f(−2)=−34 and f(−1)=3
Hence f(x) has a root in (−2,−1)
Further f′(x)=10x4+20x3+30x2+20x+10
=10x2(x2+2x+3+x2+x21)
=10x2[(x2+x21)+2(x+x1)+3]
=10x2[(x+x1)2−2+2(x+x1)+3]
=10x2[(x+x1+1)2]>0 for all x belongs to R.
f(x) is strictly increasing function. Since it is an odd degree polynomial it will have exactly one real root.
Hence, f(x) has only one real root, so ∣a∣=2.