2018 AMC 12A Problem 21: Which of the given polynomials has the greatest real root?

Question

Which of the following polynomials has the greatest real root? $\textbf{(A) } x^{19}+2018x^{11}+1 \qquad \textbf{(B) } x^{17}+2018x^{11}+1 \qquad \textbf{(C) } x^{19}+2018x^{13}+1 \qquad \textbf{(D) } x^{17}+2018x^{13}+1 \qquad \textbf{(E) } 2019x+2018$

The only root that I was able to find was the last one. I have no idea what to do from here. Does anyone have some ideas?

Answer 1

We start by observing the terms in options A to D. The leading term is degree 19 or 17 while the second term is degree 11 or 13. The numbers caught my eyes because $19-17=2$ and $13-11=2$ reminds me of factoring quadratics. I decide to compute $$A-B=x^{19}-x^{17}=x^{17}(x^{2}-1)$$ $$A-C=2018x^{11}-2018x^{13}=2018x^{11}(1-x^{2})$$ $$C-D=x^{19}-x^{17}=x^{17}(x^{2}-1)$$

Steps for the answer:

None of the options have positive real roots. The largest root then has to lie in $(-\infty,0]$. Now, when $x\in(-1,0)$, $A-B>0$, $A-C<0$, $C-D>0$, so $B>A>C>D$. Because $A$ to $D$ are all monotonously increasing on $x\in(-1,0)$ (check the derivatives), we see that $B$ has the largest real root in the interval $(-1,0)$. When $x=-\frac{2018}{2019}$, $B$ is negative while $E$ is $0$, so the root of $B$ has to be on the right of the root of $E$.

Gist:

In the figure, the green function is always larger than the red function. If in an interval, both functions are monotonously increasing and the green one is always larger, then the root of the green function is smaller. If in an interval, both functions are monotonously decreasing and the green one is always larger, then the root of the green function is larger.

Answer 2

Let us first use computer power to have the verdict, code:

polynomials = ( x^19 + 2018*x^11 + 1, 
                x^17 + 2018*x^11 + 1,
                x^19 + 2018*x^13 + 1,
                x^17 + 2018*x^13 + 1,
                2019*x + 2018 ) 
for p in polynomials:
    print ( "%s has only real root at %s"
            % ( p, p.roots(ring=AA, multiplicities=False)[0] ) )

Results:

x^19 + 2018*x^11 + 1 has only real root at -0.5006711240348649?
x^17 + 2018*x^11 + 1 has only real root at -0.500670857826262?
x^19 + 2018*x^13 + 1 has only real root at -0.556896901348879?
x^17 + 2018*x^13 + 1 has only real root at -0.5568954928592386?
2019*x + 2018 has only real root at -0.9995047052996532?

(sage was used, free software for all mathematical purposes, it can even be "infected" with non-free software, if the user possesses the one or the other compatible choice.)

Note: The above numbers are printed with a question mark at their end, but sage can does in fact exact computations with such numbers.

And "the winner is..."

x^17 + 2018*x^11 + 1

Let us prove this by human means.

• First of all, let us show that all roots are in the interval $[-1,0]$. Indeed, all five polynomials have

  • a negative value in $-1$, it is $-2018$ for the first four, and $-1$ for the last one,
  • a positive value in $0$, it is $+1$ for the first four, and bigger for the last one.

  So each polynomial has a root in the interval $[-1,0]$. The first derivative for each given polynomial is $>0$ on $\Bbb R$, so the given polynomial functions are strictly increasing, from $-\infty$ (when the argument tends to this limit) to $+\infty$ (when the argument tends to this limit). So each given polynomial has the root in the interval $[-1,0]$.

• The last polynomial has the root $-2018/2019<-9/10=-0.9$, and if we plug in this value, $0.9$, in any $P$ under the other four we get a negative number, since $0.9^{13}\approx 0.25\dots$, and $0.9^{11}\approx 0.31\dots$, so the term with the coefficient $2018$ "wins", $P(-0,9)<2018\cdot \frac 14+1<0$.

Let now $a\in(-1,0)$ be the root of $$ P=x^{17} + 2018x^{11} + 1\ ,$$ and let $$ Q=x^A + 2018x^B + 1$$ be an other polynomial in the list, so $A\in\{17,19\}$, $B\in\{11,13\}$ and either $A>17$, or $B>11$. We have $$ -1<a^{11}<a^{13}<a^{17}<a^{19}<0 \ , $$ so $$Q(a) > P(a)=0\ ,$$ so the root of $Q$ is in the interval $[-1,a]$, being thus smaller than $a$.

$\square$

---

Ths problem is solved, but it is always good to give a plot for the actors, in our case for the four polynomials, in the interval that hurts, here first $[-1,0]$.

...then $[-0.6,\ 0]$.

The following code was used:

sage: p1, p2, p3, p4 = [ plot( p, (-1,0) ) for p in polynomials[:4] ]
sage: (p1+p2+p3+p4).show()
Launched png viewer for Graphics object consisting of 4 graphics primitives
sage: p1, p2, p3, p4 = [ plot( p, (-0.6,0) ) for p in polynomials[:4] ]
sage: (p1+p2+p3+p4).show()
Launched png viewer for Graphics object consisting of 4 graphics primitives