Overview

Newton's sums are a powerful tool in algebra when trying to evaluate expressions of the form $r^n + s^n + t^n + \cdots$, where $r$, $s$, and $t$ are roots of a monic polynomial. Newton's sums can be extended to polynomials of any degree. They allow you to directly evaluate powers of sums without having to undergo heavy algebraic manipulation.

Key Ideas

• For a polynomial $x^n + a_1x^{n-1} + a_2x^{n-2} + \cdots+ a_n$, let $S_n$ be the sum of the roots to the $nth$ power
• From Newton's sums, $S_1 + a_1 = 0$, $S_2 + a_1S_1 + 2a_2 = 0$, and in general, $S_n + a_1S_{n-1} + a_2S_{n-2} + \cdots + na_n = 0$.
• Newton's sums follow a recursive pattern, so do not forget any previous terms needed to find a following term.

Core Skills

Make the Polynomial Monic

Divide by the leading coefficient so that the Newton sum recurrence applies cleanly.

Build a Recurrence

Use the relation between coefficients and power sums to compute future sums $S_n, S_{n+1}$ from previous sums $S_{n-2}, S_{n-1}$.

Combine with Symmetric Identities

Convert $r^k+s^k$ into expressions involving $r+s$ and $rs$ when a direct recurrence is shorter.

Worked Example

Let $x^2-5x+6=0$ have roots $r,s$. Find $r^3+s^3$.

We have $S_1 - 5 = 0$, so $S_1 = 5$.

From here, we have $S_2 - 5S_1 + 12 = 0$, or $S_2 = 5(5)-12 = 13$.

Now, going into our final relation, we have $S_3 - 5S_2 + 6S_1 = 0$.

Notice that we do not have a $na_n$ term in this relation since we have no other terms left.

Therefore, $S_3 = 5(13)-6(5) = 35$, which is our answer.

Notice that $x^2-5x+6=0$ can be factored as $(x-3)(x-2)$. If it is simpler to directly compute the roots and apply powers to them individually, that would be a faster solution.

However, not all polynomials have roots that can be easily computed, as seen with the next example.

More Examples

Example 1: Irrational Cubic Root Sum

If the cubic polynomial $5x^3 - 4x^2 + 3x + 2 = 0$ has roots $r, s, t$, find $r^3 + s^3 + t^3$.

This polynomial doesn't have a leading coefficient of 1, so we must make it monic.

We first must make the polynomial monic by dividing all coefficients by 5. After doing so, we get the monic cubic $x^3 - \frac{4}{5}x^2+\frac{3}{5}x + \frac{2}{5}$.

From here, we have $S_1 - \frac{4}{5} = 0$, so $S_1 = \frac{4}{5}$.

We can just keep using our recursive relation to find $S_3$.

$S_2 = \frac{4}{5}S_1-\frac{6}{5}$, so $S_2 = \frac{-14}{25}$.

Notice how the sum of the squares of the roots is negative; this indicates we have complex roots, which would be hard to find if we were to manually compute each root.

Finally, we have $S_3-\frac{4}{5}S_2+\frac{3}{5}S_1+\frac{6}{5}=0$, so $S_3 = \frac{-98}{25}$

Example 2: Artificial Polynomial

If two numbers $x,y$ follow $x+y=6$ and $x^3+y^3=100$, find $xy$.

The sums of powers of two numbers indicate that we can use Newton's sums. However, we aren't given a polynomial; to fix this, we can just make one that obeys the problem statement.

We can create $x^2-6x+a$ as our polynomial and apply Newton's sums to it.

$S_2 - 6S_1 + 2a = 0, S_2 = 36-2a$

$S_3-6S_2+aS_1=0, S_3 = 216-18a$

Setting our given values of $S_3$ equal to the one we just obtained, we have $216-18a=100$, or $a = 58/9$. Therefore, $xy = 58/9$.

Strategy Checklist

• Make the polynomial monic first.
• Compute lower powers first before higher powers.
• Avoid arithmetic mistakes; one small accidental sign flip can lead to a sum that is completely off.

Common Pitfalls

• Forgetting to convert to monic form first.
• Skipping earlier power sums needed for the recurrence.
• Lazy arithmetic mistakes.

Practice Problems