Transcript The Rational Zero Theorem
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
) =
q a n x n
+
a n-1 x n
-1 + … +
a
1
x
+
a
0 has integer coefficients and
p q p
(where is reduced) is a rational zero, then p is a factor of the constant term
a
0 and q is a factor of the leading coefficient
a n
.
EXAMPLE
:
Using the Rational Zero Theorem
List all possible rational zeros of
f
(
x
) = 15
x
3 + 14
x
2 3
x
– 2.
Solution
The constant term is –2 and the leading coefficient is 15. Possible rational zeros = Factors of the constant term, 2 Factors of the leading coefficient, 15 = = 1, 3, 2, 2 5, 15 1 3 ,
Divide
1 and
2 by
1.
2 3
Divide
1 and
2 by
3.
, 1 5 , 2 5
Divide
1 and
2 by
5.
, 1 15 , 2 15
Divide
1 and
2 by
15.
There are 16 possible rational zeros. The actual solution set to 14
x
2 3
x
– 2 = 0 is {-1,
1 / 3
,
2
f
(
x
) = 15
x
3 +
/ 5
}, which contains 3 of the 16 possible solutions.
EXAMPLE:
Solving a Polynomial Equation
Solve:
x
4
6
x
2
8
x
+ 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.
Possible rational zeros = = Factors of the constant term, 24 Factors of the leading coefficient, 1 1, 1, 2 2 3, 3, 4, 4, 1 6, 6, 8, 12, 8, 12, 24 24
EXAMPLE:
Solving a Polynomial Equation
Solve:
x
4
6
x
2
8
x
+ 24 = 0.
Solution
The graph of
f
(
x
) =
x
4
6
x
2
8
x
+ 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. 2 2 1 0 6 8 24 2 4 4 1 2 2 12 24 0
The zero remainder indicates that 2 is a root of x 4
6x 2
8x + 24
=
0.
2 8 12 1 4 6 0
The zero remainder indicates that 2 is a root of
x
3
+
2x 2
2x
12 = 0.
EXAMPLE:
Solving a Polynomial Equation
Solve:
Solution
x
4
6
x
2
8
x
+ 24 = 0.
Now we can solve the original equation as follows.
x
4 6
x
2 + 8
x
+ 24 = 0
This is the given equation.
(
x
– 2)(
x
– 2)(
x
2 + 4
x
+ 6) = 0
x
– 2 = 0 or
x
– 2 = 0 or
x
2 + 4
x
+ 6 = 0
x
= 2
x
= 2
x
2 + 4
x
+ 6 = 0
This was obtained from the first ,second synthetic division.
Set each factor equal to zero.
Solve.
EXAMPLE:
Solving a Polynomial Equation
Solve:
x
4
6
x
2
8
x
+ 24 = 0.
Solution
We can use the quadratic formula to solve
x
2 + 4
x
+ 6 = 0.
x
= = = =
b
2 2
a
4
ac
4 2 8 2 2 2
i i
2 2
We use the quadratic formula because x 2
+
4x
+
6
= 0
cannot be factored. Let a
=
1, b
=
4, and c
=
6. Multiply and subtract under the radical. Simplify.
=
2
i
2
The solution set of the original equation is { 2, 2, 2
i
i
2, 2
i
+
i
2 }
.
EXAMPLE
:
Using the Rational Zero Theorem
2
x
4 + 3
x
3 + 2
x
2 1 = 0
Solution
The constant term is –1 and the leading coefficient is 2. possible rational zeros =
Factors
.
of
.
the
.
cons
tan
t
.
term
, 1 =
Factors
.
of
.
the
.
leading
.
coefficien t
, 2 1 , 1 2
Divide
1 by
1 Divide
1 by
2
possible rational zeros 1 , 1 2 There are 4 possible rational zeros. The actual solution set to 2
x
4 is {-1,1
1 / 2
,
1 / 2
}, which contains 2 of the 4 possible solutions.
+ 3
x
3 + 2
x
2 – 1 = 0
EXAMPLE:
Solving a Polynomial Equation
2
x
4 + 3
x
3 + 2
x
2 1 = 0
Solution
The graph of
f
(
x
) 2
x
+ 3
x
+ 2
x
1 = 0 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. -1 2 3 2 0 2 1 1 -2 -1 -1 1 1 2 1 1 -1 0 2 1 1 1 2 2 2 0
The zero remainder indicates that -1 is a root of 2x 4 2x 2 -1
=
0.
+3
x
3
+
The zero remainder indicates that1 / 2 is a root of 2x 3
+
x
2
+
x -1
=
0
EXAMPLE:
Solving a Polynomial Equation
Solve:
Solution
2
x
4 + 3
x
3 + 2
x
2 1 = 0 Now we can solve the original equation as follows.
x
+1 = 0 or 2
x
4 + 3
x
3 + 2
x
2 1 = 0
This is the given equation.
(
x
+1)(
x
– )(2 2
x
2 + 2
x
+ 2) = 0
x
– 1 2 = 0 or 2
x
2 + 2
x
+ 2 = 0
This was obtained from the first ,second synthetic division.
Set each factor equal to zero.
x
= -1
x
= 1 2
x
2 +
x
+ 1 = 0
Solve.
EXAMPLE:
Solving a Polynomial Equation Solution
x
= We can use the quadratic formula to solve
x
2 +
x
+ 1 = 0.
2
b a
2 4
ac
We use the quadratic formula because x 2
+
x
+
1
= 0
cannot be factored. Let a
=
1, b
=
1, and c
=
1.
= 1 1 2 2 ( 1 ) = 1 = 2 1
i
2 3 3 4 ( 1 )( 1 )
Multiply and subtract under the radical.
The solution set of the original equation is { -1, 1 2 , 1 2 3
i
}
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
+
b i
is a root of a polynomial equation (
b
complex number
a
b i
0), then the non-real is also a root. Non-real complex roots, if they exist, occur in conjugate pairs.
Descartes' Rule of Signs
If
f
(
x
) =
a
n
x
n
coefficients.
+
a
n
1
x
n
1
+
…
+
a
2
x
2
+
a
1
x
+
a
0
be a polynomial with real
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.
EXAMPLE
:
Using Descartes’ Rule of Signs
Determine the possible number of positive and negative real zeros of
f
(
x
) =
x
3 + 2
x
2 + 5
x
+ 4.
Solution 1.
To find possibilities for positive real zeros, count the number of sign changes in the equation for
f
(
x
). Because all the terms are positive, there are no variations in sign. Thus, there are
no
positive real zeros.
2.
To find possibilities for negative real zeros, count the number of sign changes in the equation for
f
(
x
). We obtain this equation by replacing
x
with
x
in the given function.
f
(
x
) =
x
3 + 2
x
2 + 5
x
+ 4
This is the given polynomial function.
Replace x with
x.
f
(
x
) = (
x
) 3 + 2(
x
) 2 + 5
x
+ 4 =
x
3 + 2
x
2 5
x
+ 4
EXAMPLE
:
Using Descartes’ Rule of Signs
Determine the possible number of positive and negative real zeros of
f
(
x
) =
x
3 + 2
x
2 + 5
x
+ 4.
Solution
Now count the sign changes.
f
(
x
) =
x
3 + 2
x
2 5
x
+ 4
1 2 3
There are three variations in sign. # of
negative real zeros
an even integer. of
f
is either equal to 3, or is less than this number by This means that there are either 3 negative real zeros or 3 2 = 1 negative real zero.
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
+
b i
is a root of a polynomial equation (
b
complex number
a
b i
0), then the non-real is also a root. Non-real complex roots, if they exist, occur in conjugate pairs.
Complex Conjugates Theorem
Roots/Zeros that are not
Real Pairs
.
are
Complex
with an
Imaginary
component. Complex roots with Imaginary components always exist in
Conjugate
If
a + bi
(
b
≠ 0) is a zero of a polynomial function, then its Conjugate,
a - bi
,
is also a zero of the function.
Find Roots/Zeros of a Polynomial
If the known root is
imaginary
, we can use the
Complex Conjugates Theorem.
Ex: Find all the roots of If one root is
4 - i
.
f
(
x
) =
x
3 5
x
2 7
x
+ 51 Because of the Complex Conjugate Theorem, we know that
another
root must be
4 + i
.
Can the third root also be imaginary?
Example (con’t)
Ex: Find all the roots of
f
(
x
) =
x
3 5
x
2 7
x
+ 51 If one root is
4 - i
.
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)].
x
Multiply these factors:
i
x
i
=
x
2
x
4
x
4 4
i
4 +
i
=
x
2 4
x
xi
4
x
+
xi
+ 16
i
2 =
x
2 8
x
+ =
x
2 8
x
+ 17
Example (con’t)
Ex: Find all the roots of
f
(
x
) =
x
3 5
x
2 7
x
+ 51 If one root is
4 - i
.
If the product of the two non-real factors is
x
2 8
x
+ 17
x
2 8
x
: then the third factor (that gives us the neg. real root) is the quotient of
P(x)
+ 17 divided by
x
2 8
x
+ 17
x
3
x
3 5 5
x x
2 2 7 7
x x x
+ + 3 51 + 51 The third root is x = -3 0 The answer of x 3 -5x 2 -7x+ 15 =0 is {4-i,4+i,-3}
So if asked to find a polynomial that has zeros, 2 and 1 – 3i, you would know another root would be 1 + 3i. Let’s find such a polynomial by putting the roots in factor form and multiplying them together.
If
x
= the root
x
- the root then is the factor form.
x
2
x
1 3
i
x
1 + 3
i
x
2
x
1 + 3
i
x
1 3
i
x
=
x
2
x
2 2
x
2
x
3
xi
2
x
+
x
10 + 1 +
disappear when simplified.
3
i
+ 3
xi
3
i
Now multiply the x
9
i
2
-1
– 2 through
=
x
3 4
x
2
Multiply the last two factors together. All i terms should
+ 14
x
20 Here is a 3 rd degree polynomial with roots 2, 1 - 3
i
and 1 + 3
i
Conjugate Pairs
• Complex Zeros Occur in Conjugate Pairs = If
a + bi
is a zero of the function, the conjugate function must have real coefficients) the given zeros • -1, -1, 3i, -3i
a – bi
is also a zero of the function (the polynomial • EXAMPLES: Find the polynomial equation with • 2, 4 + i, 4 – i
Now write a polynomial equation of least degree that has
real coefficients
, a
leading coeff. of 1
and
1, -2+i, -2-i as zeros.
• 0= (x-1)(x-(-2+i))(x-(-2-i)) • 0= (x-1)(x+2 - i)(x+2+ i) • 0= (x-1)[(x+2) - i] [(x+2)+i] • 0= (x-1)[(x+2) 2 - i 2 ] • 0=(x-1)(x 2 + 4x + 4 Foil – (-1)) Take care of i 2 • 0= (x-1)(x 2 • 0= (x-1)(x • 0= x 3 2 + 4x 2 + 4x + 4 + 1) + 4x + 5) + 5x – x 2 Multiply – 4x – 5 • 0= x 3 + 3x 2 + x - 5
Now write a polynomial equation of least degree that has
real coefficients
, a
leading coeff. of 1
and
4, 4, 2+i as zeros.
• Note: 2+i means 2 – i is also a zero • 0= (x-4)(x-4)(x-(2+i))(x-(2-i)) • 0= (x-4)(x-4)(x-2-i)(x-2+i) • 0= (x 2 – 8x +16)[(x-2) – i][(x-2)+i] • 0= (x 2 – 8x +16)[(x-2) 2 – i 2 ] • 0= (x 2 – 8x +16)(x 2 – 4x + 4 – (– 1)) • 0= (x 2 – 8x +16)(x 2 – 4x + 5) • • 0= x 4 – 4x 3 +5x 2 – 8x 3 +32x 2 – 40x+16x 2 – 64x+80 0= x 4 -12x 3 +53x 2 -104x+80