Question

The value of the polynomial $x^8-x^5+x^2-x+1$ is:

• A. positive for all the real numbers
• B. negative for all the real numbers
• C. $0$
• D. depends on value of $x$

Answer 1

The correct option is A positive for all the real numbers

By factorisation of polynomial $p\left(x\right)=x^8-x^5+x^2-x+1$, we get:

$p\left(x\right)=x^5(x^3-1)+x(x-1)+1$

$=x(x-1)\left(x^4\right(x^2+x+1)+1)+1$

We know that, $x^4\ge 0$ and $x^2+x+1>0$ $\forall x\in R$.

Case I : $x\notin (0,1)$

As $x(x-1)\ge 0$,

$\Rightarrow p\left(x\right)\ge 1>0$ ...... (positive in range)

Case II : $[x\in (0,1\left)\right]$

Maximum value of $[-x(x-1\left)\right)]$ is $1/4$ at $x=1/2$.

And maximum value of $\left[x^4\right(x^2+x+1)+1]$ is $4$ at $x=1$ as its derivative is greater than zero in the range.

$\therefore x(x-1)\left(x^4\right(x^2+x+1)+1)>-\left(1/4\right)\times 4=(-1)$ ...... [as maxima of the terms are not coincident]

$\Rightarrow p\left(x\right)>0,x\in (0,1)$

$\therefore p\left(x\right)>0\forall x\in R$