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### Finding the Zeros of a Function

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Date: 05/12/99 at 13:38:47
From: Toni
Subject: Finding zeros

Can you tell me how I would find the zeros of
P(x) = (x^2-3)(2x-7)^2(x-1)^3

and what is meant by "state the multiplicity of each"?
```

```
Date: 05/13/99 at 17:11:53
From: Doctor Rick
Subject: Re: finding zeros

Hi, Toni.

By definition, the zeros of P(x) are the values of x such that
P(x) = 0. So finding the zeros of P(x) is the same as solving the
equation

(x^2-3)(2x-7)^2(x-1)^3 = 0

If a function is a product of expressions, P(x) = f(x)g(x)...h(x),
then P(x) = 0 if and only if one (or more) of f(x), g(x), ...h(x)
equals 0. (This is true because 0 * y = 0 for all y, and the product
of any two non-zero numbers cannot be zero.) Since your P(x) is such
a product, you need only look for zeros of the factors in the product.
The problem is equivalent to the solution of

x^2-3 = 0  or  (2x-7)^2 = 0  or  (x-1)^3 = 0

The first factor, (x^2-3), is not a product as it stands, but we can
rewrite it as a product of two linear expressions:

x^2 - 3 = (x + sqrt(3))(x - sqrt(3))

A difference of squares can always be written as the product of a sum
and a difference in this way. Since a zero of either of these factors
is a zero of (x^2 - 3), we have two new equations to solve:

x + sqrt(3) = 0  or  x - sqrt(3) = 0

Both these equations are linear - they only have the first power of
the variable. A linear equation by definition has a zero of
multiplicity 1; the zeros of these two linear equations are
x = -sqrt(3) and x = sqrt(3). These are both zeros of the whole
equation.

The second and third factors are themselves products - the second is

(2x-7)(2x-7) = 0

which is equivalent to

2x-7 = 0  or  2x-7 = 0

When x = 7/2, BOTH these equations are true. We say that x = 7/2 is a
zero of multiplicity 2 because it is a zero of two linear factors of
the function.

Likewise with the third factor,

(x-1)(x-1)(x-1) = 0

x-1 = 0  or  x-1 = 0  or  x-1 = 0

When x = 1, all 3 equations are true, so x = 1 is a zero of
multiplicity 3.

It's time to put it all together. The zeros of

P(x) = (x^2-3)(2x-7)^2(x-1)^3

are:

x = 7/2,       multiplicity 2
x = 1,         multiplicity 3
x = sqrt(3),   multiplicity 1
x = - sqrt(3), multiplicity 1

Once you get the idea, you can practically read off the zeroes and
their multiplicities directly from a function written in this form.

Here is an extra tidbit. If you were to expand P(x) as a polynomial,
the term with the highest exponent of x would be x^7 (actually 4x^7),
making this a polynomial of degree 7. The sum of the multiplicities
of the zeros of a polynomial is always the degree of the polynomial.
In this way you can be sure that you have found all the zeros.

I hope this helps.

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Equations, Graphs, Translations

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