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Zeros of Polynomials


Date: 9/24/96 at 12:14:19
From: Holly Barrow
Subject: Question

Dear Dr. Math,
     
I am a high school student and I have a question for you from a 
college algebra class:

Given that 1 is a zero of the equation P(x) = x^3-5x^2+17x-13, find 
all other zeros.
   
Thank you.


Date: 9/26/96 at 8:40:13
From: Doctor Pete
Subject: Re: Question

First, what does it mean for a polynomial to have a zero?  Well, a 
value z is a *zero* or *root* of the polynomial f(x) if f(z) = 0.  
So if 1 is a root of P(x), P(1) should be 0.  Indeed, when we plug in 
this value for x, we get:

     P(1) = 1^3 - 5(1)^2 + 17(1) - 13
          = 1 - 5 + 17 - 13
          = 0 

Now, to address your question, we will make a fundamental observation 
which relates zeros of a polynomial with its factorization. Suppose 
some polynomial f(x) has roots r[1], r[2], and r[3] (three roots in 
all), and suppose further that these are *all* the roots. Then:

     g(x) = (x-r[1])(x-r[2])(x-r[3]) = f(x)

If we evaluate g(r[1]), g(r[2]), g(r[3]), it is obvious that they are 
all 0; thus r[1], r[2], r[3] are roots of g(x). But g(x) has no other 
roots, because for g(x) = 0, at least one of the factors must be 0, 
so g = f.  This result is easily generalized to any number of roots.

How do we use this to solve our problem?  Well, it's simple.  
If

     P(x) = x^3 - 5x^2 + 17x - 13

and 1 is a root, then

     P(x) = (x-1) Q(x)

where Q(x) is a 2nd degree polynomial.  That is, (x-1) must divide 
P(x) evenly.  So let's divide P(x) by x-1:
 
                  x^2 -  4x + 13
          ______________________
     x-1 ) x^3 - 5x^2 + 17x - 13
           x^3 -  x^2
          -----------------
               - 4x^2 + 17x
               - 4x^2 +  4x
               -----------------
                        13x - 13
                        13x - 13
                        --------
                               0

Thus Q(x) = x^2 - 4x + 13, and we have:

     P(x) = (x-1)(x^2 - 4x + 13)

To find the other two roots of P(x), we now only need to solve the 
remaining quadratic Q(x).  It doesn't have integer roots (why?), 
so we use the quadratic equation, to get the roots:

     {2 + 3i, 2 - 3i}

where i is the square root of -1.

See if you can do this one (one root is x = 3/2):

     P(x) = x^3 - (7/2)x^2 - 3x + 9

If you really enjoy a challenge, try to find the roots of:

     P(x) = x^4 + 1

Hint: Assume P(x) = (x^2 + ax + 1)(x^2 + bx +1) and solve for a and b.

Doctor Pete,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Polynomials

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