Math Core

On the AIME you are rarely asked to find the roots of a polynomial. Instead you get a cubic or quartic with ugly roots and a question about some symmetric combination of them. Vieta's formulas, Newton's sums and a few evaluation tricks let you answer without ever solving the equation.

Vieta's formulas for any degree

If $P(x) = a_n x^n + a_{n-1}x^{n-1} + \dots + a_0$ has roots $r_1, r_2, \dots, r_n$ (counted with multiplicity), then

P(x) = a_n (x - r_1)(x - r_2)\cdots(x - r_n).

Expanding the right side and matching coefficients gives the elementary symmetric sums:

e_1 = \sum r_i = -\frac{a_{n-1}}{a_n}, \qquad e_2 = \sum_{i<j} r_i r_j = \frac{a_{n-2}}{a_n}, \qquad e_3 = \sum_{i<j<k} r_i r_j r_k = -\frac{a_{n-3}}{a_n},

and so on with alternating signs, ending with $r_1 r_2 \cdots r_n = (-1)^n \dfrac{a_0}{a_n}$ .

For a monic cubic $x^3 + bx^2 + cx + d$ with roots $r, s, t$ this is the familiar

r + s + t = -b, \qquad rs + st + tr = c, \qquad rst = -d.

The real power comes from a theorem: every symmetric polynomial in the roots can be written in terms of $e_1, e_2, \dots, e_n$ . So any symmetric expression is computable from the coefficients alone. Your job is to find the rewrite quickly.

Some rewrites to know cold (for three roots):

Expression	In terms of $e_1, e_2, e_3$
$r^2 + s^2 + t^2$	$e_1^2 - 2e_2$
$\dfrac{1}{r} + \dfrac{1}{s} + \dfrac{1}{t}$	$\dfrac{e_2}{e_3}$
$(r+s)(s+t)(t+r)$	$e_1 e_2 - e_3$
$r^2 s^2 + s^2 t^2 + t^2 r^2$	$e_2^2 - 2e_1 e_3$
$r^3 + s^3 + t^3$	$e_1^3 - 3e_1 e_2 + 3e_3$

For $(r+s)(s+t)(t+r)$ , notice $r + s = e_1 - t$ , so the product is $(e_1 - r)(e_1 - s)(e_1 - t)$ . That is the polynomial $(x - r)(x - s)(x - t)$ evaluated at $x = e_1$ , which leads to the next big idea.

Newton's sums

Let $p_k = r_1^k + r_2^k + \dots + r_n^k$ be the $k$ th power sum. Each root satisfies the polynomial, so for a monic cubic $x^3 - e_1x^2 + e_2x - e_3$ you have $r^3 = e_1r^2 - e_2r + e_3$ . Multiply by $r^{k-3}$ and add over all three roots:

Newton's sums for a cubic

If $r, s, t$ are the roots of $x^3 - e_1x^2 + e_2x - e_3$ and $p_k = r^k + s^k + t^k$ , then

p_1 = e_1, \qquad p_2 = e_1p_1 - 2e_2, \qquad p_3 = e_1p_2 - e_2p_1 + 3e_3,

and for $k \ge 4$ ,

p_k = e_1 p_{k-1} - e_2 p_{k-2} + e_3 p_{k-3}.

The first few formulas have the "extra" terms $2e_2$ and $3e_3$ because $p_0 = 3$ , not $1$ . From $k = 4$ on it is a clean linear recursion, and the same idea works for any degree.

Common mistake

Watch the signs. Vieta's formulas alternate: for $x^3 + bx^2 + cx + d$ , the root sum is $-b$ and the product is $-d$ . Before using Newton's sums, rewrite the polynomial as $x^3 - e_1x^2 + e_2x - e_3$ and read off $e_1, e_2, e_3$ with their signs. Also divide through by the leading coefficient first if the polynomial isn't monic.

Evaluating the polynomial

If $P(x) = a_n(x - r_1)\cdots(x - r_n)$ , then plugging in a number $c$ gives a product over the roots for free:

(c - r_1)(c - r_2)\cdots(c - r_n) = \frac{P(c)}{a_n}.

Two consequences show up constantly.

Products like $\prod (r_i^2 + 1)$ . Factor $r^2 + 1 = (r - i)(r + i)$ . Then $\prod (r_i - i)(r_i + i)$ is $P(i)P(-i)$ up to sign, and since $P$ has real coefficients, $P(-i) = \overline{P(i)}$ . So the product is $|P(i)|^2$ when $P$ is monic.
Sums like $\sum \dfrac{1}{c - r_i}$ . Take the logarithmic derivative of $P(x) = a_n \prod (x - r_i)$ :

\frac{P'(x)}{P(x)} = \frac{1}{x - r_1} + \frac{1}{x - r_2} + \dots + \frac{1}{x - r_n}.

So $\displaystyle\sum \frac{1}{c - r_i} = \frac{P'(c)}{P(c)}$ . (If you haven't seen derivatives: for a polynomial, $P'(x)$ is found by the rule that $x^k$ becomes $kx^{k-1}$ .)

Tip

When a question mentions $r + s$ , $s + t$ , $t + r$ (or "the sum of the other two roots"), replace $r + s$ by $e_1 - t$ . The expression usually becomes a function of a single root, and then an evaluation trick finishes it.

Worked examples

Worked example: Newton's sums

Let $r, s, t$ be the roots of $x^3 - 4x^2 + 2x - 7 = 0$ . Find $r^3 + s^3 + t^3$ .

Here $e_1 = 4$ , $e_2 = 2$ , $e_3 = 7$ . Then

\begin{aligned} p_1 &= 4, \\ p_2 &= e_1p_1 - 2e_2 = 16 - 4 = 12, \\ p_3 &= e_1p_2 - e_2p_1 + 3e_3 = 48 - 8 + 21 = 61. \end{aligned}

The answer is $61$ .

Worked example: Roots in geometric progression

The roots of $x^3 - 14x^2 + kx - 64 = 0$ are three real numbers in geometric progression. Find $k$ .

Write the roots as $\dfrac{m}{q}, m, mq$ . Their product is $m^3 = 64$ , so $m = 4$ is a root. Plug it in: $64 - 224 + 4k - 64 = 0$ , so $k = 56$ .

Check: the other two roots satisfy $\dfrac{4}{q} + 4q = 14 - 4 = 10$ , so $q = 2$ and the roots are $2, 4, 8$ . Indeed $2 \cdot 4 + 4 \cdot 8 + 8 \cdot 2 = 56$ .

Worked example: A product by evaluation

Let $r, s, t$ be the roots of $P(x) = x^3 - 3x^2 + 5x - 1$ . Find $(r^2 + 4)(s^2 + 4)(t^2 + 4)$ .

Since $r^2 + 4 = (r - 2i)(r + 2i)$ ,

\prod (r^2 + 4) = \prod (2i - r) \cdot \prod(-2i - r) = P(2i)\,P(-2i) = |P(2i)|^2.

(Each product of three factors $(c - r)$ equals $P(c)$ because $P$ is monic.) Compute $P(2i) = -8i + 12 + 10i - 1 = 11 + 2i$ . So the answer is $11^2 + 2^2 = 125$ .

Worked example: The logarithmic derivative

Let $r, s, t$ be the roots of $P(x) = x^3 - 2x^2 - 5x + 7$ . Find $\dfrac{1}{3 - r} + \dfrac{1}{3 - s} + \dfrac{1}{3 - t}$ .

The sum is $\dfrac{P'(3)}{P(3)}$ . Here $P(3) = 27 - 18 - 15 + 7 = 1$ and $P'(x) = 3x^2 - 4x - 5$ , so $P'(3) = 27 - 12 - 5 = 10$ . The answer is $10$ .

Without derivatives you could combine the fractions: the numerator is $\sum (3 - s)(3 - t)$ and the denominator is $P(3)$ , and you would get the same $10$ after more work.