Chapter 8 Exploring Polynomial Functions

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Chapter 8Exploring Polynomial Functions
Jennifer Huss
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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

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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 (-)
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8-1 Examples

• Determine if this expression is a polynomial in
  one variable. If it is, give the degree of the
  function.
• X2 2xy y2 This is not a function because it
  has x and y variables.
• 2a2 2a 4 This is a polynomial and its
  degree is 2.
• 12 2/n n2 This is not a polynomial because
  2/n has a negative degree.
• 34 18c4 15c6 This is a polynomial of degree 6.

• 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

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8-1 Examples (cont.)

• Decide if the graph is even or odd and tell how
  many real zeros it has.
• f(x) x3 5x 2
• f(x) x4 3x3 2

1.
2.
This is an odd function with 3 real zeros.
This is an even function with 2 real zeros.
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8-1 Problems

• Find f(3) for f(x) x5 5x4 15x2 8
• Find f(x 2) for f(x) x2 2x 5
• 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 5 3) even
function, 4 real zeros
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8-2 The Remainder and Factor Theorems

• There are two parts of the remainder theorem
• If the polynomial f(x) is divided by (x a), the
  remainder will be a number that is equal to f(a)
• I.e.. If f(x) is divided by x 4, f(4) will give
  the value of the remainder
• Dividend (quotient x divisor) remainder
• also can see this as f(x) q(x) x (x a)
  f(a)
• 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

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8-2 Examples

• 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
• 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.

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8-2 Problems

• 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.
• Given f(x) 4x2 6x 7, find f(-5) by
  synthetic division or direct substitution.
• Five the factors of x3 6x2 x 30 if one
  factors is (x 5).

1) (4x2 3x 1)(x 3) 5 2) f(-5) 63 3) (x
5), (x 2), and (x 3)
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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) lt 0 and f(b) gt 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
zero
b
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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.
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8-3 Problem

• 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.
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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

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8-4 Examples

• 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 10 3 sign changes, so 3
or 1 positive real zeros f(-x) -x3 7x2 16x
10 0 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.
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8-4 Examples (cont.)

• 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.

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8-4 Problems

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

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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.
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8-5 Problems

• List the possible rational zeros of f(x) x4
  8x3 7x 14.
• Find the rational zeros of f(x) x3 x2 8x
  12.

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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 af(x)2 bf(x) c 0
• This is a variation of ax2 bx c 0 where our
  x term could change depending on the problem

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8-6 Examples

• Solve the following equations.
• 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 0
• x2 4 x2 3 t2 6t 36
• x 4 x 3 t 6 t3 - 216
• x 2
• The solutions or zeros are This gives (t 6)(t2
  6t 36) 0.
• 2, -2, 3, and - 3. t2 6t 36 cant 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.

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8-6 Examples (cont.)

• 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.

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8-6 Problems

• Solve each equation.
• s 13 s 36 0
• x4 6x2 -8
• n3 12n2 32n 0

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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 fg(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,
  ff(x)

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8-7 Examples

• If f (1, 4) (10, 5) (6, -3) and g (5, 1)
  (4, 6)
• then find f?g.
• fg(5) f(1) 4
• fg(4) f(6) -3

2) If f(x) x 7 and g(x) x2 4, find
f?g(2) and g?f(2). f?g(2) fg(2)
g?f(2) gf(2) f(22 4)
g(2 7) f(4-4)
g(9) f(0)
(9)2 4 0 7
81 4 7 77
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8-7 Problems

• If f(x) 2x 10 and g(x) x2 1, find
  f?g(2) and g?f(2).
• If f(x) 8 2x and g(x) 3x, find fg(x).

1) f?g is 16 and g?f is 195 2) fg(x) 8
6x
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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)

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8-8 Examples

• 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) fg(x) g?f(x) gf(x)
• f 1/2(6 x) g(6
  2x)
• 6 21/2(6 x) ½6
  (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.

g(x)
f(x)
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8-8 Examples (cont.)

• 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.

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8-8 Problems

• Find the inverse of f(x) 2x 5 and graph the
  function and the inverse function.
• Determine if f(x) 3x 9 and g(x) -3x 9 are
  inverse functions.

1) f-1 (1/2)x (5/2) 2) No
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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