Indian Statistical Institute B.Math & B.Stat Solved Problems, Vinod Singh ~ Kolkata
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.
