Question

Let $P(x)=x^4+a{x}^3+b{x}^2+cx+320$ be a polynomial where $a,b,c\in R$. If $|P(4)|+\left|P^{\prime }(4)\right|+\left|P^{\prime \prime }(4)\right|\le 0$, then find the value of $11a-b-c$.

Answer 1

Answer: 9

$\Theta |P(4)|+\left|P^{\prime }(4)\right|+\left|P^{\prime \prime }(4)\right|\le 0$
$\Rightarrow P(4)=P^{\prime }(4)=P^{\prime \prime }(4)=0$
$\therefore x=4$ will be a repeated root of $P(x)=0$ repeating thrice.
Let root of $P(x)=0$ be $x_1,x_2,x_3,x_4$
$\therefore x_1=x_2=x_3=4$
P Product of roots $=x_1{x}_2{x}_3{x}_4=320$
$\Rightarrow x_4=5$
$\therefore$ Sum of roots $=4+4+4+5=-a\Rightarrow a=-17$
Product of roots taken two at a time
$=x_1{x}_2+x_1{x}_3+x_1{x}_4+x_2{x}_3+x_2{x}_4+x_3{x}_4=b$
$\Rightarrow b=16+16+20+16+20+20=108$
and product of roots taken three at a time
$=x_1{x}_2{x}_3+x_1{x}_2{x}_4+x_1{x}_3{x}_4+x_2{x}_3{x}_4=-c$
$\Rightarrow -c=64+80+80+80\Rightarrow c=-284$
$\therefore 11a-b-c=-187-108+304=$