Consider the polynomial f(x)=1+2x+3x2+4x3 Let s  be the sum of all distinct real roots of f(x)  and let t=|s| , then the real number s  lies in the interval:

The given polynomial is f(x)=1+2x+3x2+4x3 ⇒f(x)=4x3+3x2+2x+1 ⇒f'(x)=12x2+6x+2 Here a=12,b=6,c=2 ⇒  discriminant =62−4.12.2        (∵b2−4ac)                          =36−96                          =−60<0 ∴  f'(x)>0; ∀x∈R ⇒ only one real root for f(x)=0 Also f(0)=1,f(−1)=−2 ⇒ root must lie in (−1,0) Taking average of 0 and −1⇒f(−12)=14 ⇒  root must lie in (−1,−12) similarly, f(−34)=−12⇒  root must lie in (−34,−12)