Question

All the roots of the equation
$x^5+15x^4+94x^3+305x^2+507x+353=0$ are increased by some real number $k$ in order to eliminate the $4th$ degree term from the equation. Now, the coefficient of $x$ in the transformed equation is

• A. 2
• B. 1
• C. 6
• D. 0

Answer 1

Answer: (D)

Given equation is
$x^5+15x^4+94x^3+305x^2+507x+353=0$
If all roots of the equation is increased by $k$, then transformed equation is
$(x-k{)}^5+15(x-k{)}^4+94(x-k{)}^3+305(x-k{)}^2$
$+507(x-k)+353=0\dots (i)$
for eliminate 4 th degree term, the coefficient of $x^4$ is equal to zero. coefficient of $x^4=-5k+15$
So, $-5k+15=0$
$\Rightarrow k=3$
Put value in Eq. (i),
$(x-3{)}^5+15(x-3{)}^4+94(x-3{)}^3+305(x-3{)}^2$
$+507(x-3)+353=0$
$\therefore$ Coefficient of
$x={}^5{C}_4(-3{)}^4+15\left({}^4{C}_3(-3{)}^3\right)+94\left({}^3{C}_2(-3{)}^2\right)$
$+305\left({}^2{C}_1(-3)\right)+507$
$=5(81)+15(4(-27)]+94[3\times 9]+305[(2(-3)]+507$
$=405-1620+2538-1830+507=0$
$\therefore$ Coefficient of $x=0$