Overview

Polynomials appear often at AIME level. Learn how to relate roots and coefficients and how to build polynomials with given roots.

Key Ideas

• A polynomial $P(x)$ can be written as $c_nx^n+c_{n-1}x^{n-1}+\cdots+c_1x^1+c_0x^0$.
• A polynomial $P(x)$ can be written as $C\cdot (x-r_1)(x-r_2)\cdots(x-r_n)$.
• Symmetric expressions can often be rewritten in terms of a symmetric basis.

Core Skills

Key Terms (Introduction!)

For a polynomial $P$ depending on variables $x_1, x_2, \ldots, x_n$, we usually write the polynomial as $P(x_1, x_2, \ldots, x_n)$.

• The degree of a polynomial $P(x)$ is the highest exponent of a polynomial.
• A root $r$ of a polynomial $P(x)$ satisfies $P(r) = 0$.

Factor Theorem

For a polynomial $P(x)$ with root $r$, then we say $P(r) = 0$. In particular, there exists a polynomial $Q(x)$ such that $P(x) = (x-r)\cdot Q(x)$.

Two ways to interpret a polynomial

Roots. For a polynomial $P(x)$ with roots $r_1, r_2, \ldots, r_n$, we can rewrite

for a constant $C$.

Coefficients. For a polynomial $P(x)$, there exists constants $c_0, c_1, c_2, \ldots, c_n$ such that

Symmetric Sums

Definition. (Symmetric Expression) An expression $S(x_1, x_2, \ldots, x_n)$ is symmetric if for any permutation $\sigma = (\sigma_1, \sigma_2,\ldots, \sigma_n)$ of $(x_1,x_2, \ldots, x_n)$ satisfies

In general, any symmetric expression can be written in terms of a symmetric basis. The symmetric basis, in particular, is the expressions that actually occur directly due to Vieta's Theorem. The basis, in particular, is

For instance, $a^2+b^2+c^2=(a+b+c)^2 - 2(ab+bc+ca)$ and $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} = \dfrac{ab+bc+ca}{abc}$ are both symmetric expressions.

Worked Example

If $r$ and $s$ are roots of $x^2-5x+6=0$, compute $r^2+s^2$.

We have $r+s=5$ and $rs=6$. Then $r^2+s^2=(r+s)^2-2rs=25-12=13$.

More Examples (Warmup)

Example 1: Polynomial from Roots

Find the monic polynomial with roots $2$ and $-3$.

$(x-2)(x+3)=x^2+x-6$.

Example 2: Reciprocal Sum

If $r+s=4$ and $rs=3$, find $\frac{1}{r}+\frac{1}{s}$.

$\frac{r+s}{rs}=4/3$.

Example 3: Factor Theorem

If $p(x)=x^3-4x^2-x+4$, show that $x=1$ is a root and factor $p(x)$.

$p(1)=0$, so $(x-1)$ is a factor. Then $p(x)=(x-1)(x^2-3x-4)$.

Example 4: Shifting Roots! (important concept, must remember!)

Let $P(x)=x^4-8x^3+22x^2-24x+9$ have roots $r_1,r_2,r_3,r_4$. We will define $s_i=r_i-2$. Your job is to find the $\sum s_i^5$.

$P(x+2)=x^4-2x^2+1=(x^2-1)^2$, so the shifted roots are simply $1,1,-1,-1$. Hence $1^5+1^5+(-1)^5+(-1)^5=0$.

Example 5: Newton Sums (VERY IMPORTANT!)

Let $r_1,r_2,r_3,r_4$ be the roots of $x^4-3x^3+5x^2-7x+11=0$. Your job is to find $\sum r_i^5$.

We can use Newton sums to get: $p_1=3$, $p_2=-1$, $p_3=3$, $p_4=-9$, and $p_5=3p_4-5p_3+7p_2-11p_1=-82$. Thus $\sum r_i^5=-82$.

Example 6: Generalized Vieta (as a little exercise, prove that generalization!)

Let $r_1,r_2,r_3,r_4$ be the roots of $x^4+2x^3-5x^2+7x-3$. Your job is to compute $\sum_{i\ne j}\frac{r_i}{r_j}$.

$\sum_{i\ne j}\frac{r_i}{r_j}=\left(\sum r_i\right)\left(\sum\frac1{r_i}\right)-4$. By Vieta, we have that $\sum r_i=-2$ and $\sum \frac1{r_i}=\frac{\sigma_3}{\sigma_4}=\frac73$, giving us $-\frac{26}{3}$.

Example 7: A New Polynomial (harder version of example 4)

Let $P(x)=x^3-6x^2+9x-1$ have roots $r_1,r_2,r_3$. Your job is to find the polynomial whose roots are $r_i+\frac1{r_i}$.

Let $t=r+\frac1r$. We can eliminate $r$ from $r^2-tr+1=0$ and $r^3-6r^2+9r-1=0$ to get $t^3-15t^2+66t-89=0$. Thus, the polynomial is $x^3-15x^2+66x-89$.

Example 8: Reciprocal Symmetric Sums (very common!)

Let $r_1,r_2,r_3,r_4$ be the roots of $x^4-7x^3+17x^2-17x+6=0$. Your job is to compute the following: $\sum \frac1{r_i^2}$.

By Vieta, $\sum \frac1{r_i}=\frac{17}{6}$ and $\sum_{i<j}\frac1{r_ir_j}=\frac{17}{6}$. Hence, we get: $\sum \frac1{r_i^2}=\left(\frac{17}{6}\right)^2-2\left(\frac{17}{6}\right)=\frac{85}{36}$.

Example 9: Polynomial + Roots of Unity!

Let $r_1,r_2,r_3,r_4,r_5$ be the roots of $x^5-5x^4+10x^3-10x^2+5x-2=0$. Your job is to find $\sum \frac1{(r_i-1)^2}$.

Since $x^5-5x^4+10x^3-10x^2+5x-2=(x-1)^5-1$, we have $(r_i-1)^5=1$. If we let $\zeta_i=r_i-1$; then $\frac1{\zeta_i^2}=\zeta_i^3$, so the sum simply equals $\sum \zeta_i^3=0$.

More Examples (Hard):

Strategy Checklist

• Test small integer roots first.
• Use Vieta for symmetric expressions.
• Keep polynomials factored when possible.

Practice Problems