Indian Statistical Institute ( ISI ) B.Math & B.Stat : Algebra

Show that the polynomial $x^8 − x^7 + x^2 − x + 15$ has no real root. Let $f(x)=x^8 − x^7 + x^2 − x + 15$, we will show that $f(x) >0$ for all $x \in \mathbb{R}$. $f(x) = x^7(x-1)+x(x-1)+15 = (x-1)x(x^6+1)+15$ Now, $\mathbb{R} = (- \infty , 0] \cup (0,1] \cup (1, \infty)$. Note that $f(0)=f(1)=15$ When $x \in (1, \infty),$ $x,x-1 \quad and \quad x^6+1 > 0 \implies f(x) > 15 \quad \forall \quad x \in (1, \infty)$ When $x \in (- \infty , 0),$ $x \quad and \quad x-1 < 0 \quad thus \quad x(x-1) > 0$ since $x^6+1 > 0$ for any $x$, $f(x) > 15$ in this case too. When $x \in (0,1)$, $1 < x^6+1 < 2$ and $0 < x < 1$. Since both of them are positive $0 < x(x^6+1) < 2$. Further $-1 < x-1 < 0$, thus $x(x-1)(x^6+1) < 0$. Again $|x-1| < 1$ this implies $-2 < x(x-1)(x^6+1) < 0$. Thus $f(x) > 13 > 0$. Combining all the cases we see that $f(x) > 0 \quad \forall \quad x \in \mathbb{R}$ which shows $f(x)$ has no real root.