Let f(x)=2x5+5x4+10x3+10x2+10x+10 Using hit and trial method, f(−2)=−34<0 and f(−1)=3>0 Hence, f(x) has a root in (−2,−1). Again, f′(x)=10x4+20x3+30x2+20x+10 =10x2(x2+2x+3+
2
x
+
1
x2
) =10x2[(x2+
1
x2
)+2(x+
1
x
)+3] =10x2[(x+
1
x
)2+1+2(x+
1
x
)] =10x2[(x+
1
x
+1)2]>0∀x ⇒f(x) is strictly increasing function, since degree of f(x) is odd. ∴ It has exactly one real root. Therefore, f(x) has atleast one root in ⇒|a|=|−2|=2