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POLYNOMIAL FUNCTIONS
- Polynomial Division……………………………..………………
- The Rational Zero Test…………………………….…..………..
- Descarte’s Rule of Signs………………………...………………
- The Remainder Theorem…………………..…………….……..
- Finding all Zeros of a Polynomial Function…..……...………..
- Writing a Polynomial Function in Factored Form……………
- Finding the Equation of a Polynomial Function………..……..
- Even vs. Odd Functions……………………………….….……..
- Left and Right Behaviors of Polynomial Functions…….……..
- Graphing Polynomial Functions……………………….….……
- Word Problems …………………………………………………
Objectives
The following is a list of objectives for this section of the workbook.
By the time the student is finished with this section of the workbook, he/she should be able to…
- Find the quotient of a division problem involving polynomials using the polynomial long division method.
- Find the quotient of a division problem involving polynomials using the synthetic division method.
- Use the rational zero test to determine all possible rational zeros of a polynomial function.
- Use the rational zero test to determine all possible roots of a polynomial equation.
- Use Descarte’s Rule of Signs to determine the possible number of positive or negative roots of a polynomial equation.
- Find all zeros of a polynomial function.
- Use the remainder theorem to evaluate the value of functions.
- Write a polynomial in completely factored form.
- Write a polynomial as a product of factors irreducible over the reals.
- Write a polynomial as a product of factors irreducible over the rationals.
- Find the equation of a polynomial function that has the given zeros.
- Determine if a polynomial function is even, odd or neither.
- Determine the left and right behaviors of a polynomial function without graphing.
- Find the local maxima and minima of a polynomial function.
- Find all x intercepts of a polynomial function.
- Determine the maximum number of turns a given polynomial function may have.
- Graph a polynomial function.
Polynomial Division
There are two methods used to divide polynomials. This first is a traditional long division method, and the second is synthetic division. Using either of these methods will yield the correct answer to a division problem. There are restrictions, however, as to when each can be used.
Synthetic division can only be used if the divisor is a first degree binomial.
For the division problem
x x x x
, the divisor, 2 x − 1 , is a first degree binomial, so you
may use synthetic division.
There are no restrictions as to when polynomial long division may be used. The polynomial long division method may be used at any time. If the divisor is a polynomial greater than first degree, polynomial long division must be used.
The Division Algorithm
When working with division problems, it will sometimes be necessary to write the solution using The Division Algorithm.
The Division Algorithm: f ( (^) x ) = d ( (^) x ) ⋅ q ( (^) x ) + r ( x )
Simply put, the function = divisor · quotient + remainder
Is 12 divisible by 4?
Is 18 divisible by 3?
Is 15 divisible by 2?
Is 32 divisible by 8?
Based on your observations from the previous questions, what determines divisibility?
How can you determine whether or not the polynomial x^2^ − 3 x + 2 is a factor of
x^4^ + 10 x^2 − 4?
Find the quotient of each of the following. You may use synthetic or long division, but you need to know when to use each.
A)
x x x x
B)
x x x x
C)
4 2 2
x x x x
D)
x x x x x
E)
3 2 2
x x x x
F)
x x
G)
5 4 3 2 2
x x x x x x
H)
3 2 2
x x x x x
I)
x x x
The Rational Zero Test
The ultimate objective for this section of the workbook is to graph polynomial functions of degree greater than 2. The first step in accomplishing this will be to find all real zeros of the function. As previously stated, the zeros of a function are the x intercepts of the graph of that function. Also, the zeros of a function are the roots of the equation that makes up that function. You should remember, the only difference between an polynomial equation and a polynomial function is that one of them has f ( (^) x ).
You will be given a polynomial equation such as 2 x^4^ + 7 x^3^ − 4 x^2 − 27 x − 18 = 0 , and be asked to find all roots of the equation.
The Rational Zero Test states that all possible rational zeros are given by the factors of the constant over the factors of the leading coefficient.
factors of the constant = all possible rational zeros factors of the leading coefficient
Let’s find all possible rational zeros of the equation 2 x^4 + 7 x^3^ − 4 x^2 − 27 x − 18 = 0.
We begin with the equation 2 x^4 + 7 x^3 − 4 x^2 − 27 x − 18 = 0.
The constant of this equation is 18, while the leading coefficient is 2. We do not care about the (-) sign in front of the 18.
Writing out all factors of the constant over the factors of the leading coefficient gives the following. 1, 2, 3, 6, 9, 18 1, 2
These are not all possible rational zeros. To actually find them, take each number on top, and write it over each number in the bottom. If one such number occurs more than once, there is no need to write them both.
1 3 9 1, 2, 3, 6, 9, 18, , , 2 2 2
These are all possible rational zeros for this particular equation.
The order in which you write this list of numbers is not important. The rational zero test is meant to assist in the overall objective of finding all zeros to the polynomial equation 2 x^4 + 7 x^3^ − 4 x^2 − 27 x − 18 = 0. Each of these numbers is a potential root of the equation. Therefore, each will eventually be tested.
Using the rational zero test, list all possible rational zeros of the following functions.
A) f (^) ( x )= 2 x^4^ − 6 x^2 + 5 x − 15 B) f (^) ( x )= 3 x^5 − 6 x^4^ + 2 x^2 − 6 x + 12
C) f (^) ( x )= 8 x^3 − 2 x + 24 D) f (^) ( x )= 10 x^3^ − 15 x^2 − 16 x + 12
E) f (^) ( x )= − 6 x^3^ + 5 x^2 − 2 x + 18 F) f ( (^) x )= 4 x^4 − 16 x^3 + 12 x − 30
G) f ( (^) x )= 4 x^4 + 3 x^3^ − 2 x^2 + 5 x − 12 H) f ( (^) x )= x^5 − 6 x^4^ + 12 x^2 − 8 x + 36
It is important to understand, these lists of possible zeros for each of the polynomial functions above, are also lists of possible roots for the polynomial equations contained therein.
Using Descarte’s Rule of Signs, state the possible number of positive zeros for each of the following functions.
A) f (^) ( x )= 3 x^4 − 6 x^3^ + 2 x^2 − x + 2 B) f (^) ( x )= − x^5^ + 2 x^4^ − 3 x^3 − 7 x + 2
C) f (^) ( x )= 3 x^6 − 2 x^5^ + 7 x^4^ + 5 x^3 − x^2 + 2 x − 1 D) f (^) ( x )= − 6 x^4 − 5 x^2 − 8
E) f (^) ( x )= − 5 x^5^ + 6 x^4 − 3 x^2 + x − 15 F) (^) ( )^6 5 3
x 2 f = x − x + x − x + x
G) f ( (^) x )= x^5 − 3 x^4 + 2 x^3 + 4 x^2 + 5 x − 12 H) (^) ( )^4 3
x 3 f = x − x + x − x +
Using Descarte’s Rule of Signs, state the possible number of negative zeros for each of the following functions.
A) f (^) ( x )= 3 x^4 − 6 x^3^ + 2 x^2 − x + 2 B) f (^) ( x )= − x^5^ + 2 x^4^ − 3 x^3 − 7 x + 2
C) f (^) ( x )= 3 x^6 − 2 x^5^ + 7 x^4^ + 5 x^3 − x^2 + 2 x − 1 D) f (^) ( x )= − 6 x^4 − 5 x^2 − 8
E) f (^) ( x )= − 5 x^5^ + 6 x^4 − 3 x^2 + x − 15 F) (^) ( )^6 5 3
x 2 f = x − x + x − x + x
G) f ( (^) x )= x^5 − 3 x^4 + 2 x^3 + 4 x^2 + 5 x − 12 H) (^) ( )^4 3
x 3 f = x − x + x − x +
The Remainder Theorem
When trying to find all zeros of a complex polynomial function, use the rational zero test to find all possible rational zeros. Each possible rational zero should then be tested using synthetic division. If one of these numbers work, there will be no remainder to the division problem. For every potential zero that works, there may be others that do not. Are these just useless? The answer is no. Every time synthetic division is attempted, we are actually evaluating the value of the function at the given x coordinate. When there is no remainder left, a zero of the function has just been found. This zero is an x intercept for the graph of the function. If the remainder is any other number, a set of coordinates on the graph has just been found. These coordinates would aid in graphing the function.
Let P ( (^) x ) be a polynomial of positive degree n. Then for any number c ,
P ( x ) = Q ( x ) ⋅ ( x − c ) + P ( ) c ,
Where Q ( (^) x ) is a polynomial of degree n-1.
This simply means that if a polynomial P ( x ) is divided by ( x − c ) using synthetic division, the
resultant remainder is P ( ) (^) c.
When trying to find the zeros of the function f (^) ( x )= 2 x^4 + 7 x^3 − 4 x^2 − 27 x − 18 , first find all
possible rational zeros. Then evaluate each one. Here is one particular example.
2 2 7 4 27 18 0 4 6 20 14 2 3 10 7 4
In this example, (-2) is evaluated using synthetic division to see if it was a zero of the function. It turns out that (-2) is not a zero of the function, because there is a remainder of (-4).
Therefore, Using the Remainder Theorem, it can be stated that f (^) ( − (^2) ) = − 4.
You already saw that dividing by (-2) yields a result of (-4), giving us the statement f ( (^) − 2 )= − 4. This can be proven algebraically as follows.
( ) (^ )^ (^ )^ (^ )^ (^ )
( ) ( )
4 3 2 2 2 2
2 2 7 2 4 2 27 2 18 32 56 16 54 18 4
f f f
− − −
= − + − − − − − − = − − + − = −
Finding all Zeros of a Polynomial Function When solving polynomial equations, use the rational zero test to find all possible rational zeros, then use Descarte’s Rule of Signs to help narrow down the choices if possible. The fundamental theorem of Algebra plays a major role in this.
The Fundamental Theorem of Algebra Every polynomial equation of degree n with complex coefficients has n roots in the complex numbers.
In other words, if you have a 5th^ degree polynomial equation, it has 5 roots.
Example: Find all zeros of the polynomial function f (^) ( x )= 2 x^4 + 7 x^3^ − 4 x^2 − 27 x − 18.
2 x^4^ + 7 x^3^ − 4 x^2 − 27 x − 18 = 0 Begin by setting the function equal to zero. Find all possible rational zeros. 1 3 9 1, 2, 3, 6, 9, 18, , , 2 2 2
± ± ± ± ± ± ± ± ±
Once again, there are 18 possible zeros to the function. If Descarte’s Rule of Signs is used, it may or may not help narrow down the choices for synthetic division.
For this equation, there is 1 possible positive zero, and either 3 or 1 possible negative zeros.
This information was found in a previous example. Based on this, a chart may be constructed showing the possible combinations. Remember, this is a 4th^ **_degree polynomial, so each row must add up to 4.
checking each zero until a root of the equation is found..^ Now set up a synthetic division problem, and begin 1 3 0 1 1 2 2 7 4 27 18 0
1, 2, 3, 6, 9, 18, , , 2 2 2
± ± ± ± ± ± ± ± ±
-1 works as a zero of the function. There are now 3 zeros left. We can continue to test each zero, but we need to first rewrite the new polynomial. 2 x^3 + 5 x^5 − 9 x −1 8 The reason this must be done is to check using the rational zero test again. Using the rational zero test again could reduce the number of choices to work with, or the new polynomial may be factorable.
Here, we found that -3 works. The reason negative numbers are being used first here is because of the chart above. The chart says there is a greater chance of one of the negatives working rather than a positive, since there are potentially 3 negative zeros here and only one positive. Notice the new equation was used for the division.
2 x^2 − x − 6 This is now a factorable polynomial. Solve by factoring.
( 2 x + 3 )( x − 2 )= 0
x = − 3 2 and x = 2
We now have all zeros of the polynomial function. They are − 3 2 , − 1, − 3 and 2
Be aware, the remaining polynomial may not be factorable. In that case, it will be necessary to either use the quadratic formula, or complete the square.
Find all real zeros of the following functions (no complex numbers). Remember, if there is no constant with which to use the rational zero test, factor out a zero first, then proceed.
A) f (^) ( x )= x^3^ − 6 x^2 + 11 x − 6 B) f (^) ( x )= x^3^ − 9 x^2 + 27 x − 27
C) f (^) ( x )= x^3^ − 9 x^2 + 20 x − 12 D) f (^) ( x )= x^4^ − 7 x^2 + 12
E) f (^) ( x )= x^5^ − 7 x^4^ + 10 x^3 + 14 x^2 − 24 x F) f ( (^) x )= x^4 − 13 x^2 − 12 x
Writing a Polynomial Function in Factored Form
Once all zeros of a polynomial function are found, the function can be rewritten in one of several different ways.
A polynomial function may be written in one of the following ways.
- As a product of factors that are irreducible over the rationals. This means only rational numbers may be used in the factors.
- As a product of factors that are irreducible over the reals. Irrational numbers may be used as long as they are real, i.e. (^) ( x + (^3) ) ( x − (^3) ).
- In completely factored form. This may also be written as a product of
linear factors.
Complex numbers may be used in the factors, i.e. (^ x + 2 i )(^ x − 2 i ).
This section involves writing polynomials in one of the factored forms illustrated above. These are the same problems that were solved on the previous page, so there is no need to solve them again. Use the solutions previously found, to write the polynomial in the desired form.
For example, a polynomial function that has zeros of 3 and 2 ± 3 would look like the following; in completely factored form.
f (^) ( x ) = (^) ( x − (^3) ) (^) ( x − 2 + (^3) )( x − 2 − (^3) )
Notice each variable x is to the first power, so these are linear factors.
When polynomial functions are written like this, it is obvious where the x intercepts lie.
*Write the polynomial function as a product of factors that are irreducible over the reals.
A) f (^) ( x )= x^4 − 81 B) f (^) ( x )= x^4^ − 7 x^2 + 12 C) f (^) ( x )= x^3^ − x + 6
D) f (^) ( x )= x^6^ + 4 x^4^ − 41 x 2 + 36 E) f ( (^) x )= x^4^ + 10 x^2 + 9 F) f ( (^) x )= x^4 − x^3^ + 25 x^2 − 25 x
G) f ( (^) x )= x^4 − x^3^ − 2 x^2 − 4 x − 24 H) f ( (^) x )= x^4^ − x^3 − 29 x^2 − x − 30 I) f ( (^) x )= x^3 − x^2^ − 3 x + 3
J) f (^) ( x )= x^4^ − 7 x^2 + 10 K) f ( (^) x )= x^3 − 6 x^2 + 13 x − 10 L) f (^) ( x )= x^5^ + 15 x^3 − 16 x
*Write the polynomial function as a product of factors that are irreducible over the rationals.
A) f (^) ( x )= x^4 − 81 B) f (^) ( x )= x^4^ − 7 x^2 + 12 C) f (^) ( x )= x^3^ − x + 6
D) f (^) ( x )= x^6^ + 4 x^4^ − 41 x 2 + 36 E) f ( (^) x )= x^4^ + 10 x^2 + 9 F) f ( (^) x )= x^4 − x^3^ + 25 x^2 − 25 x
G) f ( (^) x )= x^4 − x^3^ − 2 x^2 − 4 x − 24 H) f ( (^) x )= x^4^ − x^3 − 29 x^2 − x − 30 I) f ( (^) x )= x^3 − x^2^ − 3 x + 3
J) f (^) ( x )= x^4^ − 7 x^2 + 10 K) f ( (^) x )= x^3 − 6 x^2 + 13 x − 10 L) f (^) ( x )= x^5^ + 15 x^3 − 16 x
Finding the Equation of a Polynomial Function
In this section we will work backwards with the roots of polynomial equations or zeros of polynomial functions. As we did with quadratics, so we will do with polynomials greater than second degree. Given the roots of an equation, work backwards to find the polynomial equation or function from whence they came. Recall the following example.
Find the equation of a parabola that has x intercepts of ( −3, 0 ) and ( 2, 0 .)
( −3, 0^ )^ and ( 2, 0 .) Given x intercepts of -3 and 2
x = − 3 x = 2 If the x intercepts are -3 and 2, then the roots of the equation are -3 and 2. Set eachroot equal to zero.
( x + 3 ) ( x − 2 ) For the first root, add 3 to both sides of the equal sign.For the second root, subtract 2 to both sides of the equal sign.
x^2^ + x − 6 Multiply the results together to find a quadratic expression.
y = x^2^ + x − 6 Set the expression equal to y, or^ f ( ) (^) x , to write as the equation of a parabola.
The exercises in this section will result in polynomials greater than second degree. Be aware, you may not be given all roots with which to work.
Consider the following example:
Find a polynomial function that has zeros of 0, 3 and 2 + 3 i. Although only three zeros are
given here, there are actually four. Since complex numbers always come in conjugate pairs, 2 − 3 i must also be a zero. Using the fundamental theorem of algebra, it can be determined that his is a 4th^ degree polynomial function.
Take the zeros of 0, 3, 2 ± 3 i , and work backwards to find the original function.
x = 0 x = 3 x = 2 ± 3 i
x = 0
x
x
2 2 2 2
2 3 2 3 4 4 9 4 4 9 4 13 0
x i x i x x i x x x x
= ± − = ± − + = − + = − − + = x ( x^ −^3 ) ( x^2^^ −^4 x +^13 )
The polynomial function with zeros of 0, 3, 2 ± 3 i , is equal to f ( (^) x ) = x x ( − (^3) ) (^) ( x^2 − 4 x + (^13) ). Multiplying
this out will yield the following.
( )
(^4 7 3 25 ) f (^) x = x − x + x − x
Find a polynomial function that has the following zeros.
A) -3, 2, 1 B) -4, 0, 1, 2 C) ±1, ± 2
D) 0, 2, 5 E) 2, 1 ± 3 F) ±4, 0, ± 2
G) -2, -1, 0, 1, 2 H) 1 ± 2, ± 3 I) 0, -
