## [Soln] 2013 USA Stanford Mathematics Tournament Algebra Qn 6

2013 USA SMT Alg 6. Compute the largest root of $x^4 - x^3 - 5x^2 + 2x + 6$.

For convenience, let $f(x) = x^4 - x^3 - 5x^2 + 2x + 6$. The first step when finding roots of polynomials is to substitute small values of $x$ into the polynomial, in the hope that one of the $\alpha$ you substitute will give $f(\alpha) = 0$. If that were the case, then $(x-\alpha)$ would be a factor of the polynomial, which will help to simplify matters.

Here are the values that we could try:

\begin{aligned} f(-3) &= 63 \\ f(-2) &= 6 \\ f(-1) &= 1 \\ f(0) &= 6 \\ f(1) &= 3 \\ f(2) &= -2 \\ f(3) &= 21 \end{aligned}

No such luck here… Also, beyond -3 and 3, the $x^4$ seems to be much bigger than the rest of the terms combined, so we are unlikely to have any other integers being roots of $f$.

Our attempt to factorise $f$ as $f(x) = (x - \alpha)g(x)$ has failed. The next best alternative would be to factorise $f$ as

$f(x) = (x^2 + ax + b)(x^2 + cx + d)$

For some values of $a,b,c,d$. Expanding the factorisation above and equating coefficients, we must have

$\begin{cases} bd &= 6 \\ bc + ad &= 2 \\ b + ac + d &= -5 \\ a + c &=-1 \end{cases}$

One could probably systematically eliminate the variables to solve for them, but it might not be straightforward as the equations above are not linear. Before doing this systematically, one possibility is to guess a solution and hope it works.

In this set-up, we could guess that $b$ and $d$ must be integers. As $bd = 6$, we should try substitituting $(b,d) = (1,6), (-1,-6), (2,3),(-2,-3)$ and hope that one of them works out. (There’s no guarantee that it will, but one can always hope!)

We are lucky: when $(b,d) = (-2, -3)$, the 4 equations above become

$\begin{cases} (-2)(-3) &= 6 \\ -2c -3a &= 2 \\ -2 + ac -3 &= -5 \\ a + c &= -1 \end{cases}$

which has the solution $(a,c) = (0,-1)$. Hence,

$f(x) = (x^2 - 2)(x^2 -x - 3),$

and so $f$‘s roots are $\pm \sqrt{2}$ and $\displaystyle\frac{1 \pm \sqrt{13}}{2}$. It is then an easy matter to compare the sizes of the roots and conclude that the largest root is $\displaystyle\frac{1 + \sqrt{13}}{2}$. QED.