Question

Let the roots of equation $f (x) = x^{5}-15x^{4}+ax^{3}+bx^{2}+cx-243=0$ 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 $f (x) = x^{5}-15x^{4}+ax^{3}+bx^{2}+cx -243=0$
$\therefore $ Sum of roots $=x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=15$
and $x_{1}. x_{2}. x_{3}. x_{4}. x_{5}=243$
$\therefore \displaystyle\sum_{i=1}^{5}x_{i}=15 \Rightarrow \underbrace{ \frac{1}{5}\displaystyle\sum_{i=1}^{5}x_{i}=3=}_{A.M.} \underbrace{ (x_{1}x_{2}x_{3}x_{4}x_{5})^{1/5} }_{G.M.}$
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 $=f (2) = (2-3)^{5} = (-1)^{5}=-1$