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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
p
If f(x) = anxn + an-1xn-1 +…+ a1x + a0 has integer coefficients and
q
p
(where is reduced) is a rational zero, then p is a factor of the constant
q
term a0 and q is a factor of the leading coefficient an.
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.
Factors of the constant term,  2
Factors of the leading coefficient, 15
1,  2
=
1,  3,  5,  15
Possible rational zeros =
= 1,  2,
 13 ,  23 ,
Divide 1
and 2
by 1.
Divide 1
and 2
by 3.
 15 ,  52 ,
Divide 1
and 2
by 5.
1 ,  2
 15
15
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.
EXAMPLE:
Solve:
Solving a Polynomial Equation
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.
Factors of the constant term, 24
Factors of the leading coefficient, 1
1,  2  3,  4,  6,  8,  12,  24
=
1
= 1,  2  3,  4,  6,  8,  12,  24
Possible rational zeros =
EXAMPLE:
Solve:
Solving a Polynomial Equation
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.
2 1 0 6 8 24
The zero remainder
indicates that 2 is a root
2
4 4 24
of x4  6x2  8x + 24 = 0.
1
2
2
12
0
2
2 8 12
1 4 6
0
The zero remainder
indicates that 2 is a root of
x3 + 2x2  2x  12 = 0.
EXAMPLE:
Solve:
Solution
Solving a Polynomial Equation
x4  6x2  8x + 24 = 0.
Now we can solve the original equation as follows.
x4  6x2 + 8x + 24 = 0
(x – 2)(x – 2)(x2 + 4x + 6) = 0
x – 2 = 0 or x – 2 = 0 or x2 + 4x + 6 = 0
x=2
x=2
x2 + 4x + 6 = 0
This is the given equation.
This was obtained from the first
,second synthetic division.
Set each factor equal to zero.
Solve.
EXAMPLE:
Solve:
Solution
Solving a Polynomial Equation
x4  6x2  8x + 24 = 0.
We can use the quadratic formula to solve x2 + 4x + 6 = 0.
b  b2  4ac
x=
2a
4  42  4 1  6 
=
2 1
4   8
=
2
4  2i 2
=
2
= 2  i 2
We use the quadratic formula because x2 + 4x + 6 = 0
cannot be factored.
Let a = 1, b = 4, and c = 6.
Multiply and subtract under the radical.
8 =
4(2)(1) = 2i 2
Simplify.
The solution set of the original equation is {2, 2,2 i i2, 2 +i i2 }.
EXAMPLE: Using the Rational Zero Theorem
2 x 4 + 3x 3 + 2 x 2  1 = 0
Solution
The constant term is –1 and the leading coefficient is 2.
Factors.of .the.cons tan t.term,1
Factors.of .the.leading.coefficient ,2
1
Divide 1
Divide 1
=
,1  2
by 1
by 2
possible rational zeros =
possible rational zeros
 1,
1
2
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.
EXAMPLE:
Solving a Polynomial Equation
2 x 4 + 3x 3 + 2 x 2  1 = 0
Solution The graph of f(x) = 2 x + 3x + 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 1 0 1
2
The zero remainder indicates
-2 -1 -1 1
4
3
that
-1
is
a
root
of
2x
+3x
+
1
2
2 1 1 -1 0
2x -1= 0.
2
1 1 1
2 2 2 0
4
3
2
The zero remainder indicates that1/2 is
a root of 2x3 +x2 + x -1= 0
EXAMPLE:
Solve:
Solution
Solving a Polynomial Equation
2 x 4 + 3x 3 + 2 x 2  1 = 0
Now we can solve the original equation as follows.
2 x 4 + 3x 3 + 2 x 2  1 = 0
This is the given equation.
1
(x +1)(x – 2 )(2x2 + 2x + 2) = 0
x +1 = 0
x = -1
or x
1
– 2=
x=
0 or 2x2 + 2x + 2 = 0
1
2
x2 + x + 1 = 0
Solve.
This was obtained from the first
,second synthetic division.
Set each factor equal to zero.
EXAMPLE:
Solution
Solving a Polynomial Equation
We can use the quadratic formula to solve x2 + x + 1 = 0.
b  b2  4ac
x=
2a
=
 1  12  4(1)(1)
We use the quadratic formula because x2 + x + 1 = 0
cannot be factored.
Let a = 1, b = 1, and c = 1.
2(1)
=
1  3
2
=
1 i 3
2
Multiply and subtract under the radical.
1
2
The solution set of the original equation is {-1, ,
 1  3i
2
}
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.
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.
EXAMPLE:
Using Descartes’ Rule of Signs
Determine the possible number of positive and negative real zeros of
f(x) = x3 + 2x2 + 5x + 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) = x3 + 2x2 + 5x + 4
Replace x with x.
f(x) = (x)3 + 2(x)2 + 5x + 4
= x3 + 2x2  5x + 4
This is the given polynomial function.
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
1
2
3
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.
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.
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.
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  5x 2  7x + 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)
3
2
Ex: Find all the roots of f (x) = x  5x  7x + 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)].
Multiply these factors:
 x   4  i    x   4 + i   = x 2  x  4 + i   x  4 + i  +  4  i  4 + i 
= x 2  4 x  xi  4 x + xi + 16  i 2
= x 2  8 x + 16  (1)
= x 2  8 x + 17
Example (con’t)
3
2
Ex: Find all the roots of f (x) = x  5x  7x + 51
If one root is 4 - i.
If the product of the two non-real factors is
x2  8x +17
then the third factor (that gives us the neg. real root) is the quotient of P(x) divided by
:
x2  8x +17
x +3
The third root
x 2  8x + 17 x 3  5x 2  7x + 51

x  5x  7x + 51
0
3
2
The answer of x3-5x2-7x+ 15 =0 is {4-i,4+i,-3}
is x = -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 then
x - the root is the factor form.
x  2x  1  3i x  1 + 3i 
x  2x 1 + 3i x 1  3i 
x  2 x
2
Multiply the last two factors
together. All i terms should
disappear when simplified.
 x  3xi  x + 1 + 3i + 3xi  3i  9i

= x  2 x  2 x + 10
2


-1
Now multiply the x – 2 through
= x  4 x + 14 x  20
3
2
2
Here is a 3rd degree polynomial with roots 2, 1 - 3i and 1 + 3i
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
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 - 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
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= (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– 4x3+5x2 – 8x3+32x2 – 40x+16x2 – 64x+80
0= x4-12x3+53x2-104x+80