Finding Polynomials with Given Zeros

To construct a polynomial from given zeros, set $x$ equal to each zero, move everything to one side, then multiply each resulting equation.

Learning Objective

Generate multiple polynomials with given zeros

Key Points

A polynomial constructed from $n$ roots will have degree $n$ or less. That is to say, if given three roots, then the highest exponential term needed will be $x^3$.
Each zero given will end up being one term of the factored polynomial. After finding all the factored terms, simply multiply them together to obtain the whole polynomial.
Because a polynomial and a polynomial multiplied by a constant have the same roots, every time a polynomial is constructed from given zeros, the general solution includes a constant, shown here as $c$.

Terms

polynomial
An expression consisting of a sum of a finite number of terms, each term being the product of a constant coefficient and one or more variables raised to a non-negative integer power, such as $a_n x^n + a_{n-1}x^{n-1} + ... + a_0 x^0$. Importantly, because all exponents are positive, it is impossible to divide by $x$.
zero
Also known as a root, a zero is an $x$ value at which the function of $x$ is equal to $0$.

Full Text

One type of problem is to generate a polynomial from given zeros. This can be solved using the property that if $x_0$ is a zero of a polynomial, then $(x-x_0)$ is a divisor of this polynomial and vice versa.

We assume that the problem statement is as follows: We are given some zeros. If it is not specified what the multiplicity of the zeros are, we want the zeros to have multiplicity one. There are no other zeros, i.e. if a number is not mentioned in the problem statement, it cannot be a zero of the polynomial we find.

Degree of the Polynomial

Remember that the degree of a polynomial, the highest exponent, dictates the maximum number of roots it can have. Thus, the degree of a polynomial with a given number of roots is equal to or greater than the number of roots that are given. If we already count multiplicity in this number, than the degree equals the number of roots. For example, if we are given two zeros, then a polynomial of second degree needs to be constructed.

Solution and Constants

If $x_1, x_2, \ldots x_n$ are the zeros of $f(x)$ and the leading coefficient of $f(x)$ is $1$, then $f(x)$ factorizes as

$f(x)=(x-x_1)(x-x_2)\cdots(x-x_n)$

This already gives us the solution of our problem: an answer to our question is just the product of all factors $(x-x_i)$, where the $x_i$ are the given zeros! However, we see that this polynomial is not unique:

For any nonzero constant $a$, we have that $(af)(x)=af(x)$ factorizes as

$af(x) = a(x-x_1)(x-x_2) \cdots (x-x_n)$

Thus if we find a solution $g(x)$ for our problem, we have actually found infinitely many solutions $cg(x)$, one for every non-zero number $c$.

Thus for given zeros $x_1, x_2, \ldots, x_n$ we find infinitely many solutions

$c(x-x_1)(x-x_2)\cdots (x-x_n)$

For example, if given $a$ and $b$ as zeros, then the resulting initial terms would be a constant $c$ times the two factors that give zeros at the appropriate place:

$c(x-a)(x-b)$

Multiplied out, this gives:

$cx^2-c(a+b)x+abc$