## Zeroes - roots of polynomials, Algebra

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We'll begin this section by defining just what a root or zero of a polynomial is.  We say that x = r is a root or zero of a polynomial, P ( x) , if P ( r )= 0 .  In other terms x= r is a root or zero of a polynomial if it is a solution to the equation P ( x)= 0 .

In the next couple of sections we will have to find all the zeroes for a given polynomial.  thus, before we get into that we eh to get some ideas out of the way about zeroes of polynomials that will help us in that procedure.

The procedure of finding the zeros of P ( x ) really amount to nothing more than solving out the equation P ( x )= 0 and already we know how to do that for second degree (quadratic) polynomials.  Hence, to help illustrate some of the ideas were going to be looking at let's get the zeroes of a couple of second degree polynomials.

Let's first determine the zeroes for P ( x)=x2+ 2x -15 .  To do this simply we solve out the following equation.

x2+ 2 x -15 =( x + 5)(x - 3)= 0  ⇒  x = -5, x = 3

Thus, this second degree polynomial has two zeroes or roots.

Now, let's determine the zeroes for P ( x)= x2 -14 x + 49 .

That will mean solving,

x2 -14x + 49 =( x - 7 )2= 0  ⇒  x = 7

Hence, this second degree polynomial contains a single zero or root. Also, remember that while we first looked at these we called a root like this a double root.

We solved out each of these by first factoring the polynomial and then by using the zero factor property onto the factored form.  While we first looked at the zero factor property we saw that it said that if the product of any two terms was zero then one of the terms had to be zero to begin with.

The zero factor property can be extended to as several terms as we need.  In other terms, if we've got a product of n terms i.e. equal to zero, then at least one of them had to be zero to start off with.  Thus, if we could factor higher degree polynomials then we could solve these as well.

81x^5-49x^3=0

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