Question

All the roots of the equation
$x^{5}+15 x^{4}+94 x^{3}+305 x^{2}+507 x+353=0$ 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
$x^{5}+15 x^{4}+94 x^{3}+305 x^{2}+507 x+353=0$
If all roots of the equation is increased by $k$, then transformed equation is
$(x-k)^{5}+15(x-k)^{4}+94(x-k)^{3}+305(x-k)^{2}$
$+507(x-k)+353=0 \,\,\,\,\,\,\ldots (i)$
for eliminate 4 th degree term, the coefficient of $x^{4}$ is equal to zero. coefficient of $x^{4}=-5 k+15$
So, $-5 k+15=0$
$ \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 $x=0$