# The Roots of a Polynomial

We can learn a lot about the roots, or the zeros, of a polynomial without knowing what the roots are. The coefficients of a polynomial tell us a lot about the roots. We will define the polynomial that we are speaking of before we begin. The polynomials we will be talking about are ones that have real coefficients and powers that are integers. Let $P(x)={{a}_{n}}{{x}^{n}}+{{a}_{n-1}}{{x}^{n-1}}+{{a}_{n-2}}{{x}^{n-2}}+$${{a}_{n-3}}{{x}^{n-3}}+\text{ }...\text{ }+$$\text{ }{{a}_{0}}$. Then, *a _{n}*,

*a*

_{n–1},

*a*

_{n–2},

*a*

_{n–3}, ...,

*a*

_{0}, are all real numbers.

- A polynomial of degree
*n*has*n*number or roots or less. The roots can be real and/or non-real. For example, a polynomial of degree 5 has at most 5 roots. How can a polynomial of degree*n*have less than*n*number of roots? Consider the polynomial $P(x)={{x}^{3}}+3{{x}^{2}}-9x+5$. This polynomial has only 2 real roots, 1 and -5, and no non-real roots. When we factor the polynomial, we see why: $P(x)=(x-1)(x-1)(x+5)$. The root 1 occurs twice. - Non-real roots of a polynomial are always conjugate pairs. If
*y*+*z*is a root, then If**i***y*-*z*is also a root.**i** - Polynomials that have an odd degree have at least 1 real root. This can easily be observed since the domain of odd functions is (–∞, +∞) and 0 falls within the domain. The function crosses the
*x*-axis at least once, which is the root of the polynomial.

## Sums of Roots

We now focus on the sums of the roots and how they are related to the coefficients. Let the polynomial $P(x)={{a}_{n}}{{x}^{n}}+{{a}_{n-1}}{{x}^{n-1}}+{{a}_{n-2}}{{x}^{n-2}}+$ ${{a}_{n-3}}{{x}^{n-3}}+\text{ }...\text{ }+\text{ }{{a}_{0}}$ be a polynomial with roots *r*_{1}, *r*_{2}, *r*_{3}, ..., *r _{n}*. Then it is possible to find $r_{1}^{k}+r_{2}^{k}+r_{3}^{k}+r_{4}^{k}+\text{ }...\text{ }+\text{ }r_{n}^{k}$, for any integer

*k*, in terms of the coefficients of the polynomial,

*a*,

_{n}*a*

_{n–1},

*a*

_{n–2},

*a*

_{n–3}, ...,

*a*

_{0}. We are going to let

*a*= 1 to make calculations easier. If it is not, we can always divide the whole polynomial with

_{n}*a*to make it 1.

_{n}If we work backward to find the polynomial coefficients, we will be able to find the coefficients in terms of the roots.

(i) $P(x)=(x-{{r}_{1}})(x-{{r}_{2}})(x-{{r}_{3}})(x-{{r}_{4}})$ $\times \ldots\times (x-{{r}_{n}})$

(ii) )$P(x)={{x}^{n}}-({{r}_{1}}+{{r}_{2}}+{{r}_{3}}+{{r}_{4}}+...+$ ${{r}_{n}}){{x}^{n-1}}+({{r}_{1}}{{r}_{2}}+$ ${{r}_{1}}{{r}_{3}}+{{r}_{1}}{{r}_{4}}+...+{{r}_{2}}{{r}_{3}}+{{r}_{2}}{{r}_{4}}+...$ $+{{r}_{n-1}}{{r}_{n}}){{x}^{n-2}}+$$({{r}_{1}}{{r}_{2}}{{r}_{3}}+{{r}_{1}}{{r}_{2}}{{r}_{4}}+$ ${{r}_{1}}{{r}_{2}}{{r}_{5}}+...+{{r}_{2}}{{r}_{3}}{{r}_{4}}+{{r}_{2}}{{r}_{3}}{{r}_{5}}+...+$ ${{r}_{n-2}}{{r}_{n-1}}{{r}_{n}}){{x}^{n-1}}+...$ $\pm{{r}_{1}}{{r}_{2}}{{r}_{3}}\times\ldots\times{{r}_{n}}$

The coefficients are:

- $a_{n}=1$
- ${{a}_{n-1}}=-({{r}_{\text{1}}}+{{r}_{\text{2}}}+{{r}_{\text{3}}}+{{r}_{\text{4}}}+\text{ }\ldots \text{ }+{{r}_{n}})$ = sum of product of roots taken 1 at a time, or simply the sum
- ${{a}_{n-2}}=\text{ }({{r}_{\text{1}}}{{r}_{\text{2}}}+{{r}_{\text{1}}}{{r}_{\text{3}}}+{{r}_{\text{1}}}{{r}_{\text{4}}}+\text{ }\ldots {{r}_{\text{1}}}{{r}_{n}})$ $\text{ }+\text{ }({{r}_{\text{2}}}{{r}_{\text{3}}}+{{r}_{\text{2}}}{{r}_{\text{4}}}+{{r}_{\text{2}}}{{r}_{\text{5}}}+\text{ }\ldots \text{ }+{{r}_{\text{2}}}{{r}_{n}})$ $\text{ }+\text{ }({{r}_{\text{3}}}{{r}_{\text{4}}}+{{r}_{\text{3}}}{{r}_{\text{5}}}+{{r}_{\text{3}}}{{r}_{\text{6}}}+\text{ }\ldots .{{r}_{\text{3}}}{{r}_{n}})\text{ }+\text{ }\ldots $ = sum of product of roots taken two at a time
- ${{a}_{n-3}}=-[({{r}_{\text{1}}}{{r}_{\text{2}}}{{r}_{\text{3}}}+{{r}_{\text{1}}}{{r}_{\text{3}}}+{{r}_{\text{1}}}{{r}_{\text{4}}}+\text{ }\ldots {{r}_{\text{1}}}{{r}_{n}})$ $\text{ }+\text{ }({{r}_{\text{2}}}{{r}_{\text{3}}}+{{r}_{\text{2}}}{{r}_{\text{4}}}+{{r}_{\text{2}}}{{r}_{\text{5}}}+\text{ }\ldots \text{ }+{{r}_{\text{2}}}{{r}_{n}})$ $\text{ }+\text{ }({{r}_{\text{3}}}{{r}_{\text{4}}}+{{r}_{\text{3}}}{{r}_{\text{5}}}+{{r}_{\text{3}}}{{r}_{\text{6}}}+\text{ }\ldots .{{r}_{\text{3}}}{{r}_{n}})\text{ }+\text{ }\ldots]$ = sum of product of roots taken 2 at a time
- ${{a}_{n-4}}=({{r}_{1}}{{r}_{2}}{{r}_{3}}+{{r}_{1}}{{r}_{2}}{{r}_{4}}+...+{{r}_{1}}{{r}_{2}}{{r}_{n}})+$ $({{r}_{1}}{{r}_{3}}{{r}_{4}}+{{r}_{1}}{{r}_{3}}{{r}_{5}}+...$ $+{{r}_{1}}{{r}_{3}}{{r}_{6}})+...+$ $({{r}_{2}}{{r}_{3}}{{r}_{4}}+{{r}_{2}}{{r}_{3}}{{r}_{5}}+...+$ ${{r}_{2}}{{r}_{3}}{{r}_{n}})+...$ = sum of product of roots taken 3 at a time
- ....
- ${{a}_{0}}=\pm ({{r}_{\text{1}}}\times {{r}_{\text{1}}}\times {{r}_{\text{1}}}\times {{r}_{\text{1}}}{{\times }_{\ldots }}\times {{r}_{n}})$ = product of the roots.
*a*_{0}is the constant.

The sign of the last term, the constant, depends on whether the polynomial is odd or even. Because the signs alternate, if the polynomial is even, then the product will have a positive sign. If it's odd, then the sign is negative.

For a polynomial of any degree, the sum of the roots is given by: $\underset{i}{\mathop{\sum }}\,{{r}_{i}}=-\frac{{{a}_{n-1}}}{{{a}_{n}}}$.

### Example 1

Let’s try it with a real polynomial. Let $P(x)=(x-1)(x+2)(x-3)$, which has roots 1, –2, and 3. Multiplying out, we have: $P(x)=({{x}^{2}}+x-2)(x-3)=$ ${{x}^{3}}-2{{x}^{2}}-5x+6$. So, *a*_{3} = 1, *a*_{2} = –2, *a*_{1} = –5, and *a*_{0} = 6.

1 – 2 + 3 = 2 = –

*a*

_{2}.

(1 × –2) + (1 × 3) + (–2 × 3) = –2 + 3 – 6 = –5 =

*a*

_{1}.

*a*

_{0}.

Remember the polynomial is odd and the sign of the last term is negative. We take the negative of

*a*

_{0}for the product.

The math checks correctly.

### Example 2

The following example involves complex roots and real roots. Let $P(x)={{x}^{4}}-5{{x}^{3}}+11{{x}^{2}}+11x-78$. There’s really no “sure-thing” method of solving for the roots for higher degrees. We can use Newton’s method as the next best thing. However, just for illustration, I derived this polynomial working forward from $P(x)=(x-2+3\mathsf{i})(x-2+3\mathsf{i})(x+2)(x-3)$. The roots are 2 + 3* i*, 2 – 3

*, –2, and 3.*

**i**2 + 3

*+ 2 – 3*

**i***– 2 + 3 = 5.*

**i**Using the formula, the sum is –

*a*

_{3}= 5.

(2 + 3

*) × (2 – 3*

**i***) × –2 × 3 = (4 + 9)(–6) = –78, which equals*

**i***a*

_{0}.

Polynomial is even so we keep the sign as is.

For the sum of the roots, only *a _{n}* and

*a*

_{n-1}are relevant. How do the other coefficients play a role? We will see their role in the sum of the square of the roots.

## Sum of Square of Roots

To find the sum of the squares, we first square the sum.

(iii) ${{({{r}_{\text{1}}}+{{r}_{\text{2}}}+{{r}_{\text{3}}}+{{r}_{\text{4}}}+\text{ }\ldots \text{ }+{{r}_{n}})}^{2}}=$ $(r_{1}^{2}+r_{2}^{2}+r_{3}^{2}+r_{4}^{2}+\text{ }\ldots \text{ }+r_{n}^{2})+$ $2({{r}_{\text{1}}}{{r}_{\text{2}}}+{{r}_{\text{1}}}{{r}_{\text{3}}}+{{r}_{\text{1}}}{{r}_{\text{4}}}+\text{ }\ldots {{r}_{\text{1}}}{{r}_{n}})$

(iv) ${{\left( \frac{{{a}_{n-1}}}{{{a}_{n}}} \right)}^{2}}=(r_{1}^{2}+r_{2}^{2}+r_{3}^{2}+r_{4}^{2}+$ $\text{ }\ldots \text{ }+r_{n}^{2})+$ $2\cdot \frac{{{a}_{n-2}}}{{{a}_{n}}}$

Solving for the squares, we have the sum of the squares of the roots:

The sum of the square of the roots is given by: $\underset{i}{\mathop{\sum }}\,r_{i}^{2}={{\left( \frac{{{a}_{n-1}}}{{{a}_{n}}} \right)}^{2}}-$ $\frac{2{{a}_{n-2}}}{{{a}_{n}}}$.

### Example 3

Using the polynomial in Example 2$P(x)={{x}^{4}}-5{{x}^{3}}+11{{x}^{2}}+11x-78$, let’s find the sum of the squares manually and using the formula. The roots are 2 + 3* i*, 2 – 3

*, –2, and 3.*

**i***)*

**i**^{2}+ (2 – 3

*)*

**i**^{2}+ (–2)

^{2}+ 3

^{2}=

4 + 12

*– 9 + 4 – 12*

**i***– 9 + 4 + 9 = 3*

**i**The formula checks.

## Sum of Higher Powers

You can find the sum of higher powers of roots on your own. The formulas for the cubes and the fourth power are below. With each higher power, it gets more and more complex.

The sum of the cube of the roots is: $\underset{i}{\mathop{\sum }}\,r_{i}^{3}=-{{\left( \frac{{{a}_{n-1}}}{{{a}_{n}}} \right)}^{3}}+\frac{3{{a}_{n-1}}{{a}_{n-2}}}{a_{n}^{2}}-$ $\frac{3{{a}_{n-3}}}{{{a}_{n}}}$.

### Example 4

We'll use the same polynomial again for the sum of the cubes of the roots manually and with the formula: $P(x)={{x}^{4}}-5{{x}^{3}}+11{{x}^{2}}+11x-78$. The roots are 2 + 3* i*, 2 – 3

*, –2, and 3.*

**i***)*

**i**^{3}+ (2 – 3

*)*

**i**^{3}+ (–2)

^{3}+ 3

^{3}=

–46 + 9

*– 46 – 9*

**i***– 8 + 27 = –73*

**i**$-{{ \frac{(-5)}{1^3} }^{3}}+\frac{3\cdot -5\cdot 11}{1}-\frac{3\cdot 11}{1}=$ $125-165-33=-73$

The formula checks.

The sum of the roots to the fourth power: $\underset{i}{\mathop{\sum }}\,r_{i}^{4}={{\left( \frac{{{a}_{n-1}}}{{{a}_{n}}} \right)}^{4}}-\frac{4a_{n-1}^{2}{{a}_{n-2}}}{a_{n}^{3}}+$ $\frac{4{{a}_{n-1}}{{a}_{n-3}}+2a_{n-2}^{2}}{a_{n}^{2}}-$ $\frac{4{{a}_{n-4}}}{{{a}_{n}}}$.

## Sum of Inverse of Roots

Now, what about the sum of the inverse of the roots? Here we use a clever tactic: divide the polynomial by the constant.

(v) $P(x)=\frac{1}{{{r}_{1}}{{r}_{2}}{{r}_{3}}\times ...\times {{r}_{n}}}{{x}^{n}}-$ $\left( \frac{1}{{{r}_{1}}{{r}_{2}}}+\frac{1}{{{r}_{1}}{{r}_{2}}}+\frac{1}{{{r}_{1}}{{r}_{3}}}+...+\frac{1}{{{r}_{n-1}}{{r}_{n}}} \right){{x}^2}+...$ $\mp \left( \frac{1}{{{r}_{1}}}+\frac{1}{{{r}_{2}}}+\frac{1}{{{r}_{3}}}+\frac{1}{{{r}_{4}}}+...+\frac{1}{{{r}_{n}}} \right)x\pm 1$.

You can see that the coefficient of the *x*^{1} term is the sum of the reciprocal of the roots. By dividing the polynomial with the product, we have essentially converted the sum of the products to the sum of the reciprocal of the products. Hence,

The sum of the reciprocal of the roots is: $\sum\limits_{i}{\frac{1}{{{r}_{i}}}}=-\frac{{{a}_{1}}}{{{a}_{0}}}$.

Note that we don’t have to worry about the sign. It will always be negative because if the polynomial is even and the constant is positive, the *x* term preceding the constant will have a negative sign. If the polynomial is odd and the constant is negative, then the *x* term preceding the constant will have a positive sign. Hence, in both cases, we have a negative sign.

### Example 5

Let’s try it with an example. Let $P(x)=(x-2)(x-5)(x+1)(x+3)(x+2)=$ ${{x}^{5}}-{{x}^{4}}-21{{x}^{3}}-11{{x}^{2}}+68x+60$. The roots are 2, 5, –1, –3, and –2. The sum is 1. Since *a*_{4} = –1, this checks. The product of the roots is –60. Since *P*(*x*) is odd, we take the negative of the constant. Hence –*a*_{0} = –60. This also checks.

Now, let’s find the sum of the reciprocals manually and with the formula.

The answer checks again.

## Last Note

Euler used the sum of the reciprocal of the roots to determine the convergence value for the sum of the inverse squares:

$\frac{{{\pi }^{2}}}{6}=\frac{1}{{{1}^{2}}}+\frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}+\frac{1}{{{4}^{2}}}+\frac{1}{{{5}^{2}}}+$ $\frac{1}{{{6}^{2}}}+\text{....}=$ $\sum\limits_{n=1}^{\infty }{\frac{1}{{{n}^{2}}}}$.

He assumed that the formula for the sum of the reciprocal of the roots works for an infinite series as well. He manipulated the Taylor series of sine to arrive with a polynomial of the form $1+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+{{a}_{3}}{{x}^{3}}+{{a}_{4}}{{x}^{4}}+{{a}_{5}}{{x}^{5}}+...$. I extended Euler's method and found the convergence of series of the form: $\sum\limits_{n=1}^{\infty }{\frac{1}{{n^2}-{{z}^{2}}}}$ and $\sum\limits_{n=1}^{\infty }{\frac{1}{{n^2}+{{z}^{2}}}}$. You can read about this topic here: Euler's Series of π.