proving no real roots exist

Question

Prove that $x^8-x^7+x^2-x+15$ has no real roots.

I did it by first assuming it has real roots and then applying Descartes rule of signs.

We find that if there are any real roots, they all must be positive numbers.

For every positive number we can prove that this expression is also positive.

Post other method for solving this.

Answer 1

Note that $$ x^8-x^7+x^2-x+15 = x^7(x-1) + x(x-1) + 15\\ = (x^7 + x)(x-1) + 15 \\ = (x^6 + 1)x(x-1) + 15 $$ so if there is to be a real root, it must be for an $x$ where $(x^6 + 1)x(x-1)$ is negative. That only happens for $x \in (0,1)$.

But if $x \in (0,1)$, then $$ |(x^6 + 1)x(x-1)| = |x^6 + 1|\cdot |x| \cdot |x-1|\\ < 2\cdot 1 \cdot 1 = 2 $$ so we have $(x^6 + 1)x(x-1) \geq -2$. But then $$ x^8-x^7+x^2-x+15 = (x^6 + 1)x(x-1) + 15 < -2+15 = 13 $$ so there cannot be any real roots. In fact, this proves that not even $x^8-x^7+x^2-x+2$ has any real roots.

Also, see this question that appeared in the "related"-column to the right.

Answer 2

It is clear that $x^8-x^7+x^2-x \ge 0$ for $x \le 0$ or $x \ge 1$.

Consider then $0 \le x \le 1$.

We have $x^8-x^7+x^2-x=x(x-1)(x^6+1) \ge -\dfrac14(x^6+1) \ge -\dfrac12$.

Therefore, $x^8-x^7+x^2-x+15 \ge 14.5 >0$ for all $x$.