The Rational Zero Theorem

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The Rational Zero Theorem
The Rational Zero Theorem gives a list of
possible rational zeros of a polynomial function.
Equivalently, the theorem gives all possible
rational roots of a polynomial equation. Not
every number in the list will be a zero of the
function, but every rational zero of the
polynomial function will appear somewhere in the
list.
The Rational Zero Theorem If f (x) anxn
an-1xn-1 a1x a0 has integer coefficients
and (where is reduced) is a rational zero,
then p is a factor of the constant term a0 and q
is a factor of the leading coefficient an.
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EXAMPLE Using the Rational Zero Theorem
List all possible rational zeros of f (x) 15x3
14x2 - 3x 2.
Solution The constant term is 2 and the
leading coefficient is 15.
Divide ?1 and ?2 by ?1.
Divide ?1 and ?2 by ?3.
Divide ?1 and ?2 by ?5.
Divide ?1 and ?2 by ?15.
There are 16 possible rational zeros. The actual
solution set to f (x) 15x3 14x2 - 3x 2 0
is -1, -1/3, 2/5, which contains 3 of the 16
possible solutions.
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EXAMPLE Solving a Polynomial Equation
Solve x4 - 6x2 - 8x 24 0.
Solution Because we are given an equation, we
will use the word "roots," rather than "zeros,"
in the solution process. We begin by listing all
possible rational roots.
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EXAMPLE Solving a Polynomial Equation
Solve x4 - 6x2 - 8x 24 0.
Solution The graph of f (x) x4 - 6x2 - 8x
24 is shown the figure below. Because the
x-intercept is 2, we will test 2 by synthetic
division and show that it is a root of the given
equation.
The zero remainder indicates that 2 is a root of
x4 - 6x2 - 8x 24 0.
2 2 8 12 1 4
6 0
The zero remainder indicates that 2 is a root of
x3 2x2 - 2x - 12 0.
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EXAMPLE Solving a Polynomial Equation
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 Solving a Polynomial Equation
Solve x4 - 6x2 - 8x 24 0.
Solution We can use the quadratic formula to
solve x2 4x 6 0.
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EXAMPLE Using the Rational Zero Theorem
Solution The constant term is 1 and the
leading coefficient is 2.
possible rational zeros
Divide ?1 by ?1
Divide ?1 by ?2
possible rational zeros
There are 4 possible rational zeros. The actual
solution set to 2x4 3x3 2x2 1 0 is
-1,1 -1/2, 1/2, which contains 2 of the 4
possible solutions.
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EXAMPLE Solving a Polynomial Equation
Solution The graph of f (x)
is shown the figure below.
Because the x-intercept is 2, we will test 2 by
synthetic division and show that it is a root of
the given equation.
The zero remainder indicates that -1 is a root of
2x4 3x3 2x2 -1 0.

The zero remainder indicates that1/2 is a root of
2x3 x2 x -1 0
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EXAMPLE Solving a Polynomial Equation
Solve
Solution Now we can solve the original
equation as follows.

This is the given equation.
x 1 0 or x 0 or 2x2 2x 2
0 Set each factor equal to zero.
x -1 x x2 x 1
0 Solve.
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EXAMPLE Solving a Polynomial Equation
Solution We can use the quadratic formula to
solve x2 x 1 0.
Let a 1, b 1, and c 1.
Multiply and subtract under the radical.
The solution set of the original equation is -1,
,
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Properties of Polynomial Equations
1. If a polynomial equation is of degree n,
then counting multiple roots separately, the
equation has n roots. 2. If a bi is a root
of a polynomial equation (b ? 0), then the
non-real complex number a - bi is also a root.
Non-real complex roots, if they exist, occur in
conjugate pairs.
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Descartes' 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 Using Descartes Rule of Signs
Determine the possible number of positive and
negative real zeros of f (x) x3 2x2 5x
4.
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EXAMPLE Using Descartes Rule of Signs
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. of
negative real zeros of f is either equal to 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.
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Properties of Polynomial Equations
1. If a polynomial equation is of degree n,
then counting multiple roots separately, the
equation has n roots. 2. If a bi is a root
of a polynomial equation (b ? 0), then the
non-real complex number a - bi is also a root.
Non-real complex roots, if they exist, occur in
conjugate pairs.
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Complex Conjugates Theorem
Roots/Zeros that are not Real are Complex with an
Imaginary component. Complex roots with
Imaginary components always exist in Conjugate
Pairs.
If a bi (b ? 0) is a zero of a polynomial
function, then its Conjugate, a - bi, is also a
zero of the function.
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Find Roots/Zeros of a Polynomial
If the known root is imaginary, we can use the
Complex Conjugates Theorem.
Because of the Complex Conjugate Theorem, we know
that another root must be 4 i. Can the third
root also be imaginary?
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Example (cont)
If one root is 4 - i, then one factor is x - (4
- i), and Another root is 4 i, another
factor is x - (4 i). Multiply these factors
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Example (cont)
The third root is x -3
The answer of x3-5x2-7x 15 0 is 4-i,4i,-3
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So if asked to find a polynomial that has zeros,
2 and 1 3i, you would know another root would
be 1 3i. Lets find such a polynomial by
putting the roots in factor form and multiplying
them together.
If x the root then x - the root is the factor
form.
Multiply the last two factors together. All i
terms should disappear when simplified.
-1
Now multiply the x 2 through
Here is a 3rd degree polynomial with roots 2, 1 -
3i and 1 3i
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Conjugate Pairs

• Complex Zeros Occur in Conjugate Pairs If a
  bi is a zero of the function, the conjugate a
  bi is also a zero of the function
  (the polynomial function must have real
  coefficients)
• EXAMPLES Find the polynomial equation with the
  given zeros
• -1, -1, 3i, -3i
• 2, 4 i, 4 i

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Now write a polynomial equation of least degree
that has real coefficients, a leading coeff. of 1
and 1, -2i, -2-i as zeros.

• 0 (x-1)(x-(-2i))(x-(-2-i))
• 0 (x-1)(x2 - i)(x2 i)
• 0 (x-1)(x2) - i (x2)i
• 0 (x-1)(x2)2 - i2 Foil
• 0(x-1)(x2 4x 4 (-1)) Take care of i2
• 0 (x-1)(x2 4x 4 1)
• 0 (x-1)(x2 4x 5) Multiply
• 0 x3 4x2 5x x2 4x 5
• 0 x3 3x2 x - 5

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Now write a polynomial equation of least degree
that has real coefficients, a leading coeff. of 1
and 4, 4, 2i as zeros.

• Note 2i means 2 i is also a zero
• 0 (x-4)(x-4)(x-(2i))(x-(2-i))
• 0 (x-4)(x-4)(x-2-i)(x-2i)
• 0 (x2 8x 16)(x-2) i(x-2)i
• 0 (x2 8x 16)(x-2)2 i2
• 0 (x2 8x 16)(x2 4x 4 ( 1))
• 0 (x2 8x 16)(x2 4x 5)
• 0 x4 4x35x2 8x332x2 40x16x2 64x80
• 0 x4-12x353x2-104x80

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