Warm

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Warm Up

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Zeros of Polynomial Functions
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The Rational Zero Theorem

• For any polynomial equation, the rational roots
  (solutions) must be such that the numerator of
  the root is a factor of the constant term of the
  polynomial and the denominator of the root is a
  factor of the coefficient of the highest degree
  term of the polynomial.
• WHAT?!

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Break Down

• To find all possible roots, we need to find the
  factors of the first coefficient and of the
  constant (the last number)
• THEN, we put the factors of the constant over the
  factors of the first coefficient and take all
  positive negative versions of that fraction.
• This will give us all POSSIBLE roots (they will
  not necessarily all work)
• To find the roots that work, we check all
  POSSIBLE roots. (Remember from yesterday if we
  divide by a number and the remainder is 0, then
  its a solution)

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Example

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Properties of Polynomial Equations

• If a polynomial equation is of degree n, the
  equation has n roots
• If the highest degree is 7, there will be 7 roots
• If the highest degree is 3, there will be 3 roots
• If abi is a root of the equation, then a-bi is
  also a root.

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Example

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-1 1 12 21 10 -1 -11 -10 1 11 10
0
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Example Cont.

• Get x211x-10
• Use the quadratic formula to find the last 2
  zeros. x-11.844 and .844
• The solutions are -1, -11.844, and .844

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Example
Solve x4 - 6x2 - 8x 24 0.
 
Start checking.
The zero remainder indicates that 2 is a root of
x4 - 6x2 - 8x 24 0.
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Example cont.
Solve x4 - 6x2 - 8x 24 0.
Solution Now we can rewrite the given
equation in factored form.
x4 - 6x2 8x
24 0 This is the given equation.
x 2 0 or x3 2x2 - 2x - 12 0
Set each factor equal to zero.
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Example cont.
Solve x4 - 6x2 - 8x 24 0.
Solution We can use the same approach to look for
rational roots of the polynomial equation x3
2x2 - 2x - 12 0, listing all possible rational
roots. However, take a second look at the figure
of the graph of x4 - 6x2 - 8x 24 0. Because
the graph turns around at 2, this means that 2 is
a root of even multiplicity. Thus, 2 must also be
a root of x3 2x2 - 2x - 12 0, confirmed by
the following synthetic division.
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Example cont.
Solve x4 - 6x2 - 8x 24 0.
Solution Now we can solve the original
equation as follows.
x4 - 6x2 8x
24 0 This is the given equation.
x 2 0 or x 2 0 or x2 4x
6 0 Set each factor equal to zero.
x 2 x 2 x2
4x 6 0 Solve.
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Example cont.
Solve x4 - 6x2 - 8x 24 0.
Solution We can use the quadratic formula to
solve x2 4x 6 0.
The solution set of the original equation is 2,
-2 - i?2, -2i?2
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What if

• What if you had a problem that when you came up
  with the possible solutions there were 15-20
  possibilities?
• Are you REALLY going to want to try all those?...
  NO, me neither! So, lets narrow it down!!

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Descartess Rule of Signs

• If f (x) anxn an-1xn-1 a2x2 a1x a0
  be a polynomial with real coefficients.
• 1. The number of positive real zeros of f is
  either equal to the number of sign changes of f
  (x) or is less than that number by an even
  integer. If there is only one variation in sign,
  there is exactly one positive real zero.
• 2. The number of negative real zeros of f is
  either equal to the number of sign changes of f
  (-x) or is less than that number by an even
  integer. If f (-x) has only one variation in
  sign, then f has exactly one negative real zero.

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Example
Determine the possible number of positive and
negative real zeros of f (x) x3 2x2 5x
4.
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Example cont.
Determine the possible number of positive and
negative real zeros of f (x) x3 2x2 5x
4.
Solution
Now count the sign changes.
f (-x) -x3 2x2 - 5x 4
There are three variations in sign. The number of
negative real zeros of f is either equal to the
number of sign changes, 3, or is less than this
number by an even integer. This means that there
are either 3 negative real zeros or 3 - 2 1
negative real zero. So, when trying possible
solutions, you can mark out ALL positive
possibilities.