Question

Let the roots of equation $f(x)=x^5-15x^4+a{x}^3+b{x}^2+cx-243=0$ are positive. If $f(x)$ is divided by $x-2$, then remainder is

• A. $0$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

Answer: (B)

Let the roots of the equation
$f(x)=0$ are $x_1,x_2,x_3,x_4$ & $x_5$
Given $f(x)=x^5-15x^4+a{x}^3+b{x}^2+cx-243=0$
$\therefore$ Sum of roots $=x_1+x_2+x_3+x_4+x_5=15$
and $x_1.x_2.x_3.x_4.x_5=243$
$\therefore \sum _{i=1}^5{x}_i=15\Rightarrow \underset{A.M.}{\underbrace{\frac{1}{5}\sum _{i=1}^5{x}_i=3=}}\underset{G.M.}{\underbrace{(x_1{x}_2{x}_3{x}_4{x}_5{)}^{1/5}}}$
Thus A.M. of $x_1,x_2,x_3,x_4$ & $x_5=G.M.$ of $x_1,x_2,x_3,x_4,x_5=3$
$\therefore x_1=x_2=x_3=x_4=x_5=3$
$\Rightarrow$ Given equation is $f(x)=(x-3{)}^5=0$ which is divided by $x-2$
$\therefore$ Remainder $=f(2)=(2-3{)}^5=(-1{)}^5=-1$