Chapter 8

Exploring Polynomial Functions

Jennifer Huss

8-1 Polynomial Functions

• The degree of a polynomial is determined by the greatest exponent when there is only one variable (x) in the polynomial

• Polynomial functions where the degree is n and a is the coefficient look like this:

• f(x) = a0xn + a1xn-1 + … +an-1x + an

• Example: f(x) = 4x4 – 3x3 + 12x2 – 7x + 2 (degree 4)

• If we know that x = 5 then f(5) replaces each x in the polynomial with a 5

• The degree of the function tells the maximum number of real zeros the function has, or the number of times the graph of the function

crosses the x-axis (ex: degree 4 function means there are at most 4 real zeros)

• A leading coefficient is the coefficient on the term of highest degree, in the example above it would be 4 because 4x4 is the term of highest degree

• See the book for more about the functions of different degrees

8-1 Polynomial Functions (cont.)

Even functions (degrees 0, 2, 4, 6, etc.) take this form:

Both sides of this graph go up (+) or both go down (-)

Odd functions (degrees 1, 3, 5, 7, etc.) take this form:

One side of this graph rises (+) and the other side falls (-)

8-1 Examples

1. Determine if this expression is a polynomial in one variable. If it is, give the degree of the function.

a) X2 + 2xy + y2 This is not a function because it has x and y variables.

b) 2a2 – 2a + 4 This is a polynomial and it’s degree is 2.

c) 12 – 2/n + n2 This is not a polynomial because 2/n has a negative degree.

d) 34 + 18c4 + 15c6 This is a polynomial of degree 6.

2. Find p(m + 2) if p(x) = 3x – 8x2 + x3.

p(m + 2) = 3(m + 2) – 8(m + 2)2 + (m + 2)3

= 3m + 6 – 8(m2 + 4m + 4) + (m + 2)(m2 + 4m + 4)

= 3m + 6 – 8m2 – 32m – 32 + m3 + 6m2 + 12m + 8

= m3 – 2m2 – 17m – 18

8-1 Examples (cont.)

3. Decide if the graph is even or odd and tell how many real zeros it has.

1. f(x) = x3 – 5x + 2

2. f(x) = x4 – 3x3 + 2

1.

This is an odd function with 3 real zeros.

2.

This is an even function with 2 real zeros.

8-1 Problems

1. Find f(3) for f(x) = x5 + 5x4 – 15x2 – 8

2. Find f(x + 2) for f(x) = x2 – 2x + 5

3. Graph f(x) = x4 – 5x2 + 4. Decide if its an even or odd function and tell how many real zeros it has.

1) 505 2) x2 + 2x + 53) even function, 4 real zeros

8-2 The Remainder and Factor Theorems

• There are two parts of the remainder theorem:1. If the polynomial f(x) is divided by (x – a), the remainder will be a

number that is equal to f(a)1. I.e.. If f(x) is divided by x – 4, f(4) will give the value of the remainder

2. Dividend = (quotient x divisor) + remainder1. also can see this as f(x) = [q(x) x (x – a)] + f(a)

2. The quotient is always a polynomial with one degree less than f(x)

• Synthetic division is helpful in solving these problems (this can also be called synthetic substitution)

• Factor theorem:

• (x – a) is a factor of f(x) if and only if the remainder (or f(a)) is equal to zero

•This is a good way to find the first factor of a polynomial

• The quotient may also be called a depressed polynomial because it has one less degree than the original polynomial

8-2 Examples

1. Use synthetic division and direct substitution to find f(4) when f(x) = x4 – 6x3 + 8x2 + 5x + 13.

4 1 -6 8 5 13 f(4) = 44 – 6(4)3 + 8(4)2 + 5(4) + 13

4 -8 0 20 OR =256 – 384 + 128 + 20 + 13

1 -2 0 5 33 f(4) = 33

2. Give the factors of x3 – 11x2 + 36x – 36 if one factor is x – 6.

6 1 -11 36 -36 So, after we divide the polynomial by x – 6

6 -30 36 we are left with x2 – 5x + 6 which we can

1 -5 6 0 solve by factoring into (x – 3)(x + 2).

This means the factors are (x – 6), (x – 3), and (x + 2).

This can also be written in the f(x) = quotient x divisor + remainder.

This would look like f(x) = (x2 – 5x + 6)(x – 6) + 0.

8-2 Problems

1. Use synthetic division to do (4x3 – 9x2 – 10x – 2) divided by (x – 3). Then write the answer in the form f(x) = [quotient x divisor] + remainder.

2. Given f(x) = 4x2 + 6x – 7, find f(-5) by synthetic division or direct substitution.

3. Five the factors of x3 + 6x2 – x – 30 if one factors is (x + 5).

1) (4x2 + 3x – 1)(x – 3) – 52) f(-5) = 633) (x + 5), (x – 2), and (x + 3)

8-3 Graphing Polynomial Functions and Approximating Zeros

• Look back at 8-1 to help with understanding finding zeros and the definition of even and odd functions

• Location Principle:– If y = f(x) is a polynomial function and you have a and b such that

f(a) < 0 and f(b) > 0 then there will be some number in between a and b that is a zero of the function

• A relative maximum is the highest point between two zeros and a relative minimum is the lowest point between two zeros

a

b

zero

8-3 Example

Graph the function f(x) = -2x3 – 5x2 + 3x + 2 and approximate the real zeros.

There are zeros at approximately -2.9, -0.4, and -0.8.

8-3 Problem

1. Graph f(x) = x3 + x2 – 4x – 4 and approximate the real zeros. Show the relative minimum and maximum on the graph.

1) The real zeros are approximately -2, -1, and 2.

8-4 Roots and Zeros

• The Fundamental Theorem of Algebra says that every polynomial equation has at least one root in the set of complex numbers

• Another way to state it: a polynomial with degree n has exactly n roots in the set of complex numbers

• Remember: roots can be imaginary (complex numbers)• The Complex Conjugates Theorem says that if a + bi is a zero of a

polynomial function then a – bi is also a zero of the function• Descartes’ Rule of Signs says that if f(x) is a polynomial with its

terms arranged in order of decreasing power (ex: x3, x2, x) then:• The number of positive real zeros is given by the number of sign changes of

the coefficients of f(x), or less than the number of sign changes by an even number

• The number of negative real zeros is given by the number of sign changes of the coefficients of f(-x), or less than the number of sign changes by an even number

• Ex: 5 sign changes for f(x) means 5, 3, or 1 positive real zeros

8-4 Examples

1. Give the possible number of positive real zeros, negative real zeros, and imaginary zeros of f(x) = x3 – 7x2 + 16x – 10. Then find all the zeros if one zero is 3 – i.

f(x) = x3 – 7x2 + 16x – 103 sign changes, so 3 or 1 positive real zeros

f(-x) = -x3 – 7x2 – 16x – 100 sign changes, so no negative real zeros

Since the degree is 3 on this polynomial we should have 3 zeros. If we have 3 positive real zeros there will be no imaginary zeros. If we have 1 positive real zero there will be 2 imaginary zeros.

So, 3 positive real zeros or 1 positive real zero and 2 imaginary zeros.

Since 3 – i is one zero, 3 + i will also be a zero.

f(x) = [x – (3 – i)][x – (3 + i)](?)

f(x) = [ x2 – (3 – i)x – (3 + i)x + (3 – i)(3 + i)](?)

f(x) = (x2 – 3x + xi – 3x – xi + 9 – i2) (?)

f(x) = (x2 – 6x + 10) (?)

So now we need to find the (?), which is the third factor, by long division. x – 1 x2 – 6x + 10 ) x3 – 7x2 + 16x – 10

-(x3 – 6x2 + 10x) _

-x2 + 6x – 10

-(-x2 + 6x – 10)

0

So, (x – 1) is the third factor, which means the third zero is 1.

The zeros are 3 + i, 3 – i, and 1.

8-4 Examples (cont.)

2. Given that 1 and 1 + i are two zeros of a polynomial, write the polynomial of the least degree having these zeros.

If 1 + i is a zero, 1 – i is another zero.

f(x) = [x – (1 + i)] [x – (1 – i)] (x – 1)

f(x) = [ x2 – (1 + i)x – (1 – i)x + (1 + i)(1 – i)] (x – 1)

f(x) = [x2 – x – xi – x + xi + 1 – i2] (x – 1)

f(x) = (x2 – 2x + 2) (x – 1)

f(x) = x3 – 2 x2 + 2x – x2 + 2x – 2

f(x) = x3 – 3x2 + 4x – 2

The polynomial is x3 – 3x2 + 4x – 2.

8-4 Problems

1. State the number of positive real zeros, negative real zeros, and imaginary zeros in f(x) = 16x3 + 6x2 – 7x + 3.

2. Given f(x) = x3 + 6x + 20 and one of its zeros as 1 – 3i, find all of the zeros of this function.

1) 2 or 0 positive real zeros, 1 negative real zero, 2 or 0 imaginary zeros.2) The zeros are 1 – 3i, 1 + 3i, and -2.

8-5 Rational Zero Theorem

• The rational zero theorem helps us find zeros when we have large numbers that are hard to factor

• Rational Zero Theorem says that if you have a polynomial f(x) = a0xn + …+ an-1x + an, then you can find zeros by doing p divided by q if p is a factor of an and q is a factor of ao

• A similar theorem, the Integral Zero Theorem, says that if a0 = 1 and an= 0, then q = 1 which makes p/q= p. This means that all the zeros of this function will simply be the factors of an.

• To find which zeros actually work, you need to do the Descartes’ Rule of Signs and graph the function

8-5 Example

List the possible rational zeros for f(x) = 3x4 – 2x3 – 5. Then graph the function to see which are the actual rational zeros.

a0 = 3 which means q = 1, 3

an = -5 which means p = 1, 5

Possible rational zeros are: 1 , 5 , 1, 5 or 1, 5, 1/3, 5/3 1 1 3 3

The real zeros are -1 and 5/3.

8-5 Problems

1. List the possible rational zeros of f(x) = x4 – 8x3 + 7x – 14.

2. Find the rational zeros of f(x) = x3 – x2 – 8x + 12.

1) 1, 2, 7, 42) The zeros are 2 and -3.

8-6 Using Quadratic Techniques to Solve Polynomial Equations

• Sometimes we want to solve or factor a polynomial that is not degree 2 (x2)

• We try to force the polynomial into the quadratic form so then we can factor and solve it

• The quadratic form is: a[f(x)]2 + b[f(x)] + c = 0• This is a variation of ax2 + bx + c = 0 where our “x” term could

change depending on the problem

8-6 Examples

Solve the following equations.

1) x4 – 7x2 + 12 = 0 2) t3 – 216 = 0(x2)2 – 7(x2) + 12 = 0 First, you must look at the graph to find the (x2 – 4)(x2 – 3)= 0 first zero at x = 6. Then perform long division.

x2 – 4 = 0 x2 – 3 = 0x2 = 4 x2 = 3 t2 + 6t + 36x = 4 x = 3 t – 6 t3 - 216x = 2The solutions or zeros are This gives (t – 6)(t2 + 6t + 36) = 0. 2, -2, 3, and - 3. t2 + 6t + 36 can’t be factored so we use

the quadratic formula.

t = - 6 (6)2 – 4(1)(36) = -3 3i 3 2(1)

The zeros are 6, -3 + 3i 3, and -3 – 3i 3.

8-6 Examples (cont.)

3. y – 8 y + 7 = 0

( y)2 – 8( y) + 7 = 0

( y – 7)( y – 1) = 0

y – 7 = 0 y – 1 = 0

y = 7 y = 1

y = 49 y = 1

The solutions or zeros are 1 and 49.

8-6 Problems

Solve each equation.

1. s – 13 s + 36 = 0

2. x4 – 6x2 = -8

3. n3 + 12n2 + 32n = 0

1) 16 and 812) 2, -2, 2,and - 23) 0, -4, and -8

8-7 Composition of Functions

• The composition of functions is when you combine two functions to create one multi-step function

• The composition function f ○ g needs to have the range of g as part of the domain of f (the output of g is part of the input for f)

• The composition f ○ g is written as f[g(x)]

• In these problems you solve g(x) to get some value a, and then you solve f(a) to get the final answer

• Two functions may not have a composition if we find g(x) to be a, but f(a) is not possible

• Iteration is a special composition where the function combines with itself, for example, f[f(x)]

8-7 Examples

1) If f = {(1, 4) (10, 5) (6, -3)} and g = {(5, 1) (4, 6)}

then find f○g.

f[g(5)] = f(1) = 4

f[g(4)] = f(6) = -3

f○g

5 4

1 6

1 10 6

4 5 -3

Domain (x’s) of g

Range (y’s) of g

Domain (x’s) of f

Range (y’s) of f

2) If f(x) = x + 7 and g(x) = x2 – 4, find [f○g](2) and [g○f](2).

[f○g](2) = f[g(2)] [g○f](2) = g[f(2)]

= f(22 – 4) = g(2 + 7)

= f(4-4) = g(9)

= f(0) = (9)2 – 4

= 0 + 7 = 81 – 4

= 7 = 77

8-7 Problems

1. If f(x) = 2x + 10 and g(x) = x2 – 1, find [f○g](2) and [g○f](2).

2. If f(x) = 8 – 2x and g(x) = 3x, find f[g(x)].

1) [f○g] is 16 and [g○f] is 1952) f[g(x)] = 8 – 6x

8-8 Inverse Functions and Relations

• Two functions, f and g, are inverse functions (opposites) if their composition gives the identity function (x)

• [f○g](x) = x and [g○f](x) = x

• To check for inverses, take both compositions and see if both equal x

• Also, if you graph the functions the inverse functions should be mirror images or reflections of one another across the line y = x

• f-1 mean “f inverse” and f = g-1 means f is the inverse of g

• If f and f-1 are inverse functions, f(a) = b and f-1(b) = a• This means that the ordered pair (a, b) will change to (b, a) for the inverse function

• To write an inverse function, switch the x and the y of the equation• y = ax + b changes to x = ay + b

• Inverse relations means that a relation (set of ordered pairs) can be changed into an inverse by switching (a, b) to (b, a)

8-8 Examples

1. Determine whether f(x) = 6 – 2x and g(x) = ½(6 – x) are inverse functions. Check by graphing.

In order to determine this we will find [f○g](x) and [g○f](x).

[f○g](x) = f[g(x)] [g○f](x) = g[f(x)]

= f [1/2(6 – x)] = g(6 – 2x)

= 6 – 2[1/2(6 – x)] = ½[6 – (6 – 2x)]

= 6 – 6 + x = ½ (6 – 6 + 2x)

= x = ½ (2x) Yes, f(x) and g(x)

= x are mirror images.

Yes, they are inverse functions since both compositions equal x

and the graphs are mirror images.

f(x)g(x)

8-8 Examples (cont.)

2. Find the inverse of f(x) = x + 3. Then graph both functions to verify they are inverses.

To find the inverse, switch y and x.f(x) = x + 3 y = x + 3 x = y + 3 y = x – 3 f-1 = x – 3 The graphs are mirror images across y = x.

Check:[f ○f-1](x) = f(x – 3) [f-1○f](x) = f-1(x + 3)

= (x – 3) + 3 = (x + 3) – 3 = x = x

Yes, f-1 = x – 3 is the inverse function.

8-8 Problems

1. Find the inverse of f(x) = 2x + 5 and graph the function and the inverse function.

2. Determine if f(x) = 3x – 9 and g(x) = -3x + 9 are inverse functions.

1) f-1 = (1/2)x – (5/2) 2) No

8-8B Square Root Functions and Relations

• Square root functions can never be negative if we want to find answers that are real numbers

• The square root graph looks like the following:

• For examples and practice problems, see the textbook

y = x