Question

Roots of the equation $f\left(x\right)=x^6-12x^5+b{x}^4+c{x}^3+d{x}^2+ex+64=0$ are positive. Remainder when $f\left(x\right)$ is divided by $x-1$ is

• A. $2$
• B. $1$
• C. $3$
• D. $10$

Answer 1

The correct option is B $1$

Given

$x^6-12x^5+b{x}^5+c{x}^3+d{x}^2+ex+64=0$

Since ${\alpha }_{1,}{\alpha }_2,{\alpha }_3,{\alpha }_4,{\alpha }_5,{\alpha }_6$ are the roots of the equation

Let us consider the sum of the roots

${\alpha }_1,{\alpha }_1,+.......+{\alpha }_6=12\to 1$

The product of roots would be

${\alpha }_1,{\alpha }_2,.......{\alpha }_6=64\to 2$

If we apply arithmetic and geometric mean to equation roots

$A.M>G.M$

${\alpha }_1+{\alpha }_2+.......+{\alpha }_6\ge ({\alpha }_1,{\alpha }_2.......+{\alpha }_6{)}^{{1/6}^{}}$

$\frac{12}{6}\ge ({\alpha }_1,{\alpha }_2,.......{\alpha )}^{1/6}$

$2^6\ge \frac{12}{6}\ge ({\alpha }_1,{\alpha }_2,.......{\alpha }_6)$

${\alpha }_1,{\alpha }_2,.......{\alpha }_6)\le 64$

$({\alpha }_1,{\alpha }_2,.......{\alpha }_6)=64$

The roots product to equality if all the roots are equal

$\Rightarrow ({\alpha }_1={\alpha }_2=.......={\alpha }_6)$

Since the root equation is 2

$(x-2)=0$ defines the root of the equation.

If we divide $(x-2)$ by $(x-1)$ the remainder is $1$.