TS EAMCET 5 May 2018 Shift 1 Solved Paper

The given equation is,
x5+15x4+94x3+305x2 +507x+353=0
If all the roots is increased by k, then
(x−k)5+15(x−k)4+94(x−k)3 +305(x−k)2 +507(x−k)+353=0 …… (1)
The value of k is calculated as,
x4=−5k+15
−5k+15=0
k=3
Substitute the value in equation (1),
(x−3)5+15(x−3)4+94(x−3)3 +305(x−3)2+507(x−3)+353=0
The coefficient of x is,
x=[5C4(−3)4+15(4C3(−3)3) +94 (3C2(−3)2)+305(2C1(−3))+507]
=5(81)+15(4)(−27)+94(3)(9) +305(2)(3)+507
=405−1620+2538−1830 +507
=0