# How to show that a polynomial does not have real roots

How to show generally that a polynomial does not have real roots. Well, for eg lets take the polynomial $x^8-x^7+x^2-x+15$ . Here the power($n=8$) is even so it can have real roots or it might not have real roots.

Something which I thought was to find the minima and show that if the minima of $p(x)$ is greater than $0$ and $a_1$ that is the coefficient of $x^8$ are both greater than $0$ then we cannot have real roots . But in this case the derivative is $8x^7-7x^6+2x-1$ and I cannot find minima for it . So what should I do in this example . Well it is already given this polynomial does not have real roots , but I have to prove it.

Also even if I get that this does not have any real roots then is this a general method for all kinds of polynomials ?

Edit: I know Strum's theorem is one general way to solve such questions but this question is from an undergrad entrance paper and I guess a method under the reach of calculus or something similar will suffice better.

• general method is Sturm's theorem – Michael Galuza Jul 26 '15 at 5:43
• But this is a question in an undergrad entrance paper and strum's theorem has not been taught yet . So is there any method under the reach of calculus @MichaelGaluza – user210387 Jul 26 '15 at 5:45
• If real root exists then it must be in $(0, 1)$ – Michael Galuza Jul 26 '15 at 5:50
• For the specific polynomial you've given, we can write it as $x(x-1)(x^6+1)+15 = 0$. So any real root $x$ must satisfy $x(x-1)(x^6+1) = -15 < 0$, so we must have $0 < x < 1$. However, in that range, $|x| < 1$, $|x-1| < 1$, and $|x^6+1| < 2$. Thus, $|x(x-1)(x^6+1)| < 2 < 15$, a contradiction. Of course this isn't a general method. – JimmyK4542 Jul 26 '15 at 5:50
• Perhaps a more general way is to express as a sum of squares. Here the polynomial is the same as $$\frac57 \left(2 x^2+x-\frac{23}{10}\right)^2+7 \left(\frac{x^3}2+\frac{x^2-3 x-5}7\right)^2+\left(x^4-\frac{x^3}2-x^2-x-1\right)^2+(2 x-1)^2+\frac{113}{20}$$ – Macavity Jul 26 '15 at 12:52

Clearly there is no negative root as all terms are positive for $x < 0$. The question remains if there are positive roots. Here is a simple way which often works.

Case 1: $0 < x <1$.

$$P(x) = (15-x) + (x^2-x^7) + x^8 > 0$$ as each term is positive.

Case 2: $x > 1$. Similarly $$P(x) = (x^8-x^7) + (x^2-x) + 15 > 0$$

as $x=1$ is not a root, we are done.

• (A minor remark: The constant term is nonzero, so that $0$ is also not a root.) – Benjamin Dickman Jul 26 '15 at 11:58
• @BenjaminDickman Quite right. I figured it is obvious, but perhaps in the interest of being complete it should also be noted. – Nemo Jul 26 '15 at 12:57

The polynomial can be rewritten $$x^7(x-1) + x(x-1) + 15.$$

Unless $x$ is between $0$ and $1$, the first two terms are positive, and so the polynomial is positive.

Even if $x$ is between $0$ and $1$, the first two terms are tiny in magnitude, certainly each individually greater than $-1$, so that when $15$ is added to their sum, the result is positive.

Thus the polynomial has no real roots.

It should be clear that on the interval $[-1,1]$ you have $|x^8-x^7+x^2-x|\leq |x^8|+|x^7|+|x^2|+|x|\leq 4$ and so $x^8-x^7+x^2-x+15\geq 11$

Further you should notice that $x^8-x^7>0$ when $|x|>1$ and that $x^2-x>0$ when $|x|>1$, so $x^8-x^7+x^2-x+15\geq 11$ for all $x$

After thinking over night, it occurred to me that the poster's method could be made to work, with a little effort... \begin{align} f(x) &= x^8 − x^7 + x^2 − x + 15 \\ f'(x) &= 8x^7 - 7x^6 + 2x - 1 \\ f''(x) &= 56 x^6 - 42 x^5 + 2 \\ f'''(x) &= 336 x^5 -210 x^4 \end{align}

The roots of $f'''(x)$ are $\{0,0,0,0,\frac{105}{168}\}$. For $x<0$ and $x \in \left(0,\frac{105}{168}\right)$, $f'''(x) < 0$ (check at $1/2$, getting $\sim 10 - 13$). For $x > \frac{105}{168}$, $f'''(x)>0$ (check at $1$). This means $f''(x)$ is monotonically nonincreasing on $\left(-\infty, \frac{105}{168}\right)$ and monotonically increasing on $\left( \frac{105}{168}, \infty\right)$. So if $f''$ has any roots, it has at most one in each of those intervals and may have one at $\frac{105}{168}$.

$f''\left(\frac{105}{168}\right) \approx (56 \times 0.6 - 42)(0.6)^5 + 2 \approx -6^6/10^5 + 2$, but $6^6 = 36 \times 36 \times 6 \approx 6000$, so $f''\left(\frac{105}{168}\right) \in (1,2)$. (The exact result is $2-\frac{6675.72 \dots}{10^5} = \frac{43661}{32768}$, so alter the estimate to make one happy.) Consequently, $f''$ has no roots and we discover $f'$ is monotonically increasing.

$f'(0) = -1$ and $f'(1) = 2$, so $f'$ has a root in $(0,1)$ and $f$ has a global minimum there.

On $(0,1)$, $x^7 > x^8$ and $x > x^2$, so both $x^8 - x^7$ and $x^2 - x$ lie in $(-1,0)$. But then a lower bound for $f$ on $(0,1)$ is $-1+-1+15 > 0$, so $f$ is positive on all of $\mathbb{R}$. (In fact, the minimum is $14.7454\!\dots$ occurring when $x = 0.530791\!\dots\,$.)

Therefore, $f$ has no real roots.

(It would have been nice if the four roots of $f'''$ at zero hadn't all fled the real line while passing to second derivative; I think handling that is the only "nuts and bolts" technique this example didn't use.)