The document explains the process of finding a polynomial function based on given roots and their multiplicities. It details the steps of identifying zeros, determining behaviors based on multiplicity, adjusting for the y-intercept, and ensuring the correct end behavior of the polynomial. The final polynomial derived from the example is f(x) = -1/10(x + 5)(x + 1)(x - 2)².

1. Finding the polynomial from data
Tyler Murphy
September 2, 2014
The

2. rst thing to realise is that this is essentially the reverse process of sketching a graph.
Consider the following problem:
1 Example:
Find the polynomial function with f(5) = f(1) = f(2) = 0 where 2 has an even
multiplicity and 5 and -1 have an odd multiplicity. Additionally, f(0) = 2, f(x) 0
for all x 5 and f(x) 0 for all x 2.
1.1 solution:
1.1.1 Step 1: The Zeroes
First consider what your zeroes are and what that looks like in terms of a polynomial
function.
In this case, the zeroes are -5, -1,and 2.
But what does this mean? Consider the function f(x) = x 5: What would the zero
be here? What about the function f(x) = (x 3)(x 5) What are the zeroes here?
This should give you an idea about the pattern involved here. If your zero is x = a
then your polynomial has a term (x a).
So this function has the terms (x 5); (x 1); and (x 2).
1.1.2 Using the Multiplicity
First, you have to recall what the multiplicity of a function tells us. If the multiplicity of a
zero is even, then the graph simply touches down at that point. If the multiplicity is odd,
then the graph passes through that zero.
1

3. So for this problem, we have an odd multiplicity at -5 and -1. So we know the graph
passes through these points. We also know that the function only touches at x = 2 instead
of passing through it because the multiplicity of 2 is even.
So we know that our polynomial now looks something like this:
f(x) = (x + 5)1(x + 1)1(x 2)2
Think about why this makes sense. Since -5 is the solution to x + 5 = 0 and it has an
odd multiplicity, we know that the polynomial must have a term that looks like (x + 5)k
where k is some odd number. This follows for the other terms.
But what if the multiplicity of -5 is some other odd number than 1? What if it's a
large odd number like 101? It very well could be for another polynomial that hits these
same points. Know that there are in