# Polynomial Functions - DOC

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```					                      Polynomial Functions
Note: It is assumed that you are familiar with long division or synthetic division of
polynomials.

Examples of polynomial functions are f ( x )  4x 3  3x 2  7 x  3 and
and g( x )  5x 4  2x  1 .
The topic of this study guide is to find the x-intercepts also known as rational zeros of
polynomial functions, and to be able to make a graph of these functions - finding the x-
intercepts is essential to making the graph. Let us first look at some general aspects of
polynomial functions. Typically they look as follows

far right

local max

far left           local min

or maybe like

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far left                                  far right

local max

local min

local min

The local maxima and minima are sometimes collectively known as the turning points of
the function. The following statements are considered basic knowledge concerning
polynomial functions:
I)     A function of degree n can have at most n x-intercepts. Thus f ( x )  2x 3  4x  5 ,
which is of degree 3, can have at most 3 x-intercepts.
II)    A function of degree n can have at most (n-1) turning points. This means that the
function just mentioned can have at most 3-1=2 turning points.
III)   The far left and far right behavior of a graph is either up or down. This means that
for "large" x values, say x = 20, the graph will continue to go steeply up or down.
Likewise for x values on the far left like x = -20. The turning points will not be to
the far left or right but around the origin. The far right and far left behavior of a
polynomial can easily be determined by calculating f(20) and f(-20). If f(20) is
positive the function goes up on the far right, if it is negative the function goes
down on the far right. The same holds true for the far left. Thus the function
mentioned earlier goes up to the far right since f(20)>0 and down to the far left
since f(-20)<0.

Let us turn our attention to the x-intercepts or real zeros of a polynomial function: if a
function can be completely factored into binomials then it'll be easy to find the x-
intercepts. Just set f(x)=0, factor the function, and use the zero product property to find
the real zeros. If the function cannot be factored then we will have to look for other ways
to find all the x intercepts. Let us first consider a few 'factorable' functions.

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Example 1 : f ( x)  2 x 3  3 x 2  18 x  27
Set f(x)  0 to find the x  intercepts :
0  x 2 (2 x  3)  9(2 x  3)
0  (2 x  3( x 2  9)
0  (2 x  3)( x  3)( x  3)

The zeros are -3/2, -3 and 3. If you talk about zeros this notation is good enough, if you
are calling them x-intercepts you should present the answer as (-3/2,0), (-3,0) and (3,0).

Example 2 : g ( x )  x 4  25x 2  144
Set f x   0 to find the real zeros :
0  x 4  25x 2  144
0  ( x 2  16)(x 2  9)
0  x  4)(x  4 ( x  3)(x  3)

so that the zeros are ±4 and ±3.

However, the vast majority of polynomial functions will not easily be factorable if at all.
There are a few techniques that are useful in such cases; amongst them are the Factor
Theorem and the Rational Zero Theorem.

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The Factor Theorem

The factor theorem states that for any x=c, if f(c)=0 then (x-c) is a factor of f(x), and vice
versa. As an example consider f ( x )  x 3  2x 2  5x  6 . Clearly, no factoring technique
discussed thus far is able to bring relief here. The factor theorem, however, gets you
started. Try an easy x-value, such as x=1 and you will find that f(1)=0 so that, according
to the factor theorem (x-1) is a factor of f(x). Thus, the division of f(x) by (x-1) should
have a remainder of zero. Performing the division ( x 3  2x 2  5x  6)  ( x  1) , either by
long division or by synthetic division, will give you the reduced polynomial x 2  x  6 so
we can say that x 3  2x 2  5x  6  ( x  1)( x 2  x  6) and after factoring the trinomial
we'll have x 3  2x 2  5x  6  ( x  1)( x  3)( x  2) . Thus the real zeros of the function
are -2, 1 and 3.
After plotting the x- and y-intercepts and looking at the far left and far right behavior we
can now easily make the graph. All that we need to do still is to take a few simple x-
values and insert them into f(x) in order to get a few extra points plotted and somewhat
get a more accurate graph. It looks like this:

As a second example try to find the x-intercepts of g ( x )  x 3  5x  2 . The usual
factoring techniques yield nothing but the factoring theorem helps us out. Try a few easy
x-values like x=1 or x=2 and you will find that g(1)≠0 so (x-1) is not a factor of g(x) but
since g(2)=0 then (x-2) is a factor of g(x). Performing the division ( x 3  5x  2)  ( x  2) ,
one way or the other, you will get the reduced polynomial x 2  2x  1 and so
g( x )  ( x  2)( x 2  2x  1) . The trinomial can not be factored but you can nevertheless
find the x-intercepts by setting g(x)=0. Using the zero product property means that x-2=0
or that x 2  2x  1  0 . This last equation can be solved with the quadratic formula and

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gives you x  1  2 . The x-intercepts of g(x) are thus (2,0)
(1  2 ,0) and (1  2 ,0) . As you see, the factor theorem is a very powerful and new
factoring technique. The graph of this function can also be made after plotting all the
intercepts and consider the far left and far right behavior. It looks like this:

You may now think that it could take a long time before you finally find an x-value such
that f(x)=0. This could have been true 20 years ago but ever since pc technology has been
introduced to make our lives easier it is a different situation. Go to Microsoft Excel, open
a new workbook and in column A enter values like -5,-4,-3,-2,-1,0,1,2,3,4,5 or so. Of
course you make use of the fill handle (the little black box on the right hand bottom side
of a selected cell). Type in only the first three values, select them, put the pointer on the
fill handle and see the shape of the pointer changing and then just drag the fill handle
downwards. You will see the other values appearing without you doing any typing. Next,
go to cell B1 and type (=A1^3-5*A1+2) and press the enter key. Drag down this formula
again in column B using the fill handle and see a zero appearing next to the x-value 2.
What do you do if none of the real zeros is a whole number? You can't conceivably try all
possible fractions and plug them into the factor theorem. You don't have to. The rational
zero theorem will help us out here.

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The Rational Zero Theorem

Look at the name of this theorem. What does rational mean? A rational number is another
word for fraction. Thus, this theorem helps you only to find rational zeros that are
fractions or can be written as fractions. Since it can be shown that something like 2 can
not be expressed as a fraction this theorem will not help you find such rational zeros.
The Rational Zero Theorem states that a list of all possible rational zeros of a polynomial
function can be made be listing all values of the quantity (p/q) where p is a factor of the
last coefficient in the polynomial and q is a factor of the first coefficient in the
polynomial. To illustrate this look at P( x )  2x 3  9x 2  2x  9 . The last coefficient is -9
and has factors ±1, ±3 and ±9. These are the possible values for p. The first coefficient is
2 and has factors ±1 and ±2 which are as the values for q. The quotient (p/q) then has the
values of ±1, ±1/2, ±3, ±3/2, ±9 and ±9/2. This is then the list of the possible values for
the rational zeros of the function. You can now try each of these values to see if its
function value equals zero but this is time consuming. Excel helps us out again. First
select the first 20 cells or so in column A and format them so they can contain fractions.
With the cells selected, go for Format, Cells, Fraction, up to two digits and press OK.
Now you can enter the list of possible zeros one by one: like 1,-1, ½, -1/2 etc. In cell B1,
enter the function as (=2*A1^3+9*A1^2-2*A1-9) and use the fill handle again to drag
down this function. As a result you will see 0's next to 1, -1 and -9/2. Since a polynomial
of degree 3 can have up to 3 real zeros you have just found all of them. Of course you
could also have done this through factoring since P(x)  (x  1)(x  1)(2x  9) . The graph
is as follows:

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The next example can not be done by factoring.
Consider Q( x )  4x 4  35 x 3  71x 2  4x  6 . The values for p are ±1, ±2, ±3 and ±6.
The values for q are ±1, ±2 and ±4 so that the list of all possible rational zeros is ±1, ±1/2,
±1/4, ±2, ±3, ±3/2, ±3/4 and ±6. Using Excel in the way mentioned in this guide you find
that 3 and -1/4 are zeros. Since there may be as many as 4 real zeros in this function, the
search goes on. According to the factor theorem both (x-3) and (4x+1) - try not to use
(x+1/4) - are factors of the polynomial. Thus
4x 4  35 x 3  71x 2  4x  6  ( x  3)( 4x  1)(some polynomial ) . This can be rewritten as
4x 4  35 x 3  71x 2  4x  6  (4x 2  11x  3)(some polynomial ) and division gives us
that this polynomial is x 2  6x  2 . Thus, Q(x) factors as
4x 4  35 x 3  71x 2  4x  6  ( x  3)( 4x  1)( x 2  6x  2) . In order to find the x-
intercepts we only have to set Q(x)=0 and get that x-3=0 or 4x+1=0 or x 2  6x  2  0 .
This last equation can be solved using the quadratic formula and gives the zeros
x  3  7 . Thus, the x-intercepts are (3,0), (-1/4,0) and (3  7 ,0) . After looking at the
far left and far right behavior, the maximum number of turning point it may have and
taking a few nice x values to improve accuracy you can try to draw the graph. It is a nasty
exercise however due to the wild numbers in this problem. Try it in Winplot. Good luck!
Basically we know the shape however if we take all the factors mentioned above into
consideration.

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Descartes' Rule of Signs

This is a detail maybe worth mentioning, and maybe not. The rule of signs nowadays
only serves as an extra check to see if you have correctly found the zeros.
The rule of signs says that the number of positive real zero's of polynomial function is
equal to the number of variations in sign between the subsequent terms of the function or
equal to that number decreased by an even integer. Look at the last function we
considered. In Q(x) the sign changes going from the first to the second term, then again
between the second and third term and also between the fourth and the fifth term. There
are three changes in sign so we expect either 3 or 3-2=1 positive real zeros. We found
three of them.
The rule of signs furthermore says that the number of negative real zero is equal to the
number of variations in sign between the subsequent terms of the Q(-x) or equal to that
number decreased by an even integer. In our case, replacing x by (–x) in the function we
get: Q( x )  45 ( x ) 4  35 ( x ) 3  7( x ) 2  4( x )  6  45 x 4  35 x 3  7 x 2  4x  6 . The
number of changes in sign is only one. Thus we know that we cannot expect to find more
negative real zeros than the one we found.

Time for a minor detail in this guide - a bit of terminology. Suppose you have a function
like f ( x )  (3x  1)( 2x  5) 2 ( x  1) 3 . The zeros are obviously 1/3, 5/2 and -1. Because the
5/2 really appears twice, it is called a zero with multiplicity 2 just as "-1" is called a zero
with multiplicity 3. The total number of zeros counting the multiplicity of each zero will
be 6, which is the degree of this polynomial.

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Zeros in general
What if you do not confine yourself to real zeros but say just zeros? In that case you also
need to list the imaginary and complex numbers that make the function equal to zero.
Look at the function h ( x )  2x 3  x 2  8x  12 which factors as h ( x )  (2x  3)( x 2  4) .
If you limit yourself only to real zeros you'll just have 3/2. If you do not, then you have to
go on factoring to get h(x)  (2x  3(x  2i)(x  2i) and find zeros of 3/2, -2i and 2i. Of
course you can also use the quadratic formula to find these values since x 2  4  0 is a
Or consider the function L( x )  x 3  x  2 . Since L(1)=0, the factor theorem says that
(x-1) is a factor of L(x). The division ( x 3  x  2)  ( x  1) gives x 2  x  2 so that
L( x )  ( x  1)( x 2  x  2) . In order to find the zeros you set L(x)=0 and find that x-1=0
1 i 7
or x 2  x  2  0 . The quadratic formula gives us x            so that you now have
2
found the three zeros. The fundamental theorem of algebra says that a polynomial
function of degree n will have exactly n zeros, counting multiplicity. There may be up to
n real zeros.

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