Transcript Real Zeros of Polynomial Functions

Real Zeros of
Polynomial
Functions
Section 2.3
Objectives
• Use long division to divide polynomials by other
polynomials.
• Use synthetic division to divide polynomials by
binomials of the form (x-k).
• Use the Remainder and Factor Theorems.
• Use the Rational Zero Test to determine possible
rational zeros of polynomial functions.
• Use Descarte’s Rule of Signs and the Upper and
Lower Bound Rules to find zeros of polynomials.
Dividing Polynomials
Dividing a polynomial by a polynomial
uses a “long division” technique that
is similar to the process known as
long division in dividing two numbers,
which is reviewed on the next slide.
Dividing Polynomials
168
43 7256
43
295
258
376
344
32
Divide 43 into 72.
Multiply 1 times 43.
Subtract 43 from 72.
Bring down 5.
Divide 43 into 295.
Multiply 6 times 43.
Subtract 258 from 295.
Bring down 6.
Divide 43 into 376.
Multiply 8 times 43.
We then write our result as
Subtract 344 from 376.
32
168 .
Nothing to bring
43
down.
Dividing Polynomials
As you can see from the previous
example, there is a pattern in the long
division technique.
Divide
Multiply
Subtract
Bring down
Then repeat these steps until you
can’t bring down or divide any longer.
We will incorporate this same repeated
technique with dividing polynomials.
Division of Polynomials
Dividing Polynomials
Long division of polynomials is similar to long division of
whole numbers.
When you divide two polynomials you can check the
answer using the following:
dividend = (quotient • divisor) + remainder
The result is written in the form:
dividend  divisor 
quotient +
remainder
divisor
Long Division of Polynomials
• Procedure:
1. Arrange the terms of both polynomials in descending
order of the exponents of the variable.
2. If either the dividend or the divisor has missing terms,
insert these terms with coefficients of 0.
3. Divide
4. Check, by multiplying the divisor times the quotient
and adding the remainder.
•
If the remainder is zero, then the divisor is a
factor of the dividend.
Example: Divide x2 + 3x – 2 by x – 1 and check the answer.
x + 2
x  1 x 2  3x  2
x2 +
x
2
x
1. x x   x
x
2. x( x  1)  x 2  x
2
2x – 2
3. ( x 2  3x)  ( x 2  x)  2 x
2x + 2
–4
4. x 2 x  2 x  2
x
5. 2( x  1)  2 x  2
remainder
Answer: x + 2 +
6. (2 x  2)  (2 x  2)  4
–4
x 1
Check: (x + 2) (x + 1) + (– 4) = x2 + 3x – 2
quotient divisor remainder
dividend
correct
Example: Divide 4x + 2x3 – 1 by 2x – 2 and check the answer.
x2 + x
+ 3
2x  2 2x  0x  4x  1
3
2
2x3 – 2x2
2x2
Since there is no x2 term in the
dividend, add 0x2 as a placeholder.
+ 4x
2x2 – 2x
6x – 1
6x – 6
5
Answer:
x2
+x+3

Write the terms of the dividend in
descending order.
5
2x  2
Check: (x2 + x + 3)(2x – 2) + 5
= 4x + 2x3 – 1
3
2
x
1.
2. x 2 (2 x  2)  2 x 3  2 x 2
 x2
2x
2
2
x
3
3
2
2
3. 2 x  (2 x  2 x )  2 x
4.
x
2x
5. x(2 x  2)  2 x 2  2 x
6.(2 x 2  4 x)  (2 x 2  2 x)  6 x
8. 3(2 x  2)  6 x  6
7. 6 x  3
2x
9.(6 x  1)  (6 x  6)  5  remainder
Example: Divide x2 – 5x + 6 by x – 2.
x – 3
x  2 x 2  5x  6
x2 – 2x
– 3x
+6
– 3x + 6
0
Answer: x – 3 with no remainder.
Check: (x – 2)(x – 3) = x2 – 5x + 6
Example: Divide x3 + 3x2 – 2x + 2 by x + 3 and check the answer.
x2 + 0x
– 2
x  3 x 3  3x 2  2 x  2
x3 + 3x2
Note: the first subtraction
eliminated two terms
from the dividend.
0x2
Answer:
–2+
+ 2
– 2x – 6
Therefore, the quotient
skips a term.
x2
– 2x
8
8
x3
Check: (x + 3)(x2 – 2) + 8
= x3 + 3x2 – 2x + 2
Your Turn:
2
x
 3x  2 by x  1 .
Divide
x +2
x  1 x2  3x  2
- ( x2 + x)
2x + 2
- (2x + 2)
0
CHECK:
(x + 1)(x + 2)
= x2 + 2x + x + 2
= x2 + 3x + 2
Your Turn:
Divide x 4  x2  1
by x  1 .
3
x3 - x2 + 2x - 2

x  1 x4  0x3  x2  0x  1 x  1
- ( x4 + x3)
-x3 + x2
- (-x3 – x2)
2x2 + 0x
- ( 2x2 + 2x)
-2x + 1
- ( -2 x - 2)
3
3
3
2
x  x  2x  2 
x 1
Polynomial Long Division:
Example:
2x4 + 3x3 + 5x – 1
x2 – 2x + 2
2x2
x  2 x  2 2 x  3x  0 x  5 x  1
2
4
2x4 = 2x2
x2
3
2
2x2
+7x
+10
x  2 x  2 2 x  3x  0 x  5 x  1
2
4
-(
2x4
3
2
-4x3 +4x2
)
7x3 - 4x2 +5x
-( 7x3 - 14x2 +14x
7x3 = 7x
x2
10x2 - 9x
)
-1
-( 10x2 - 20x +20 )
11x - 21
remainder
The solution is written:
11x  21
2 x  7 x  10  2
x  2x  2
2
• Quotient + Remainder over Divisor
Your Turn:
• y4 + 2y2 – y + 5
y2 – y + 1
• Answer: y2 + y + 2 +
3
y2 – y + 1
Your Turn:
(5x3 – 4x2 + 7x  2) ÷ (x2 + 1).
Solution:
5x  4
2
3
2
x  0x  1 5x  4x  7x  2
5x 3  0x 2  5x
 4x  2x  2
2
Quotient is 5x – 4
with a remainder
of 2x + 2.
(5x3
–
4x2
+ 7x  2) ÷
(x2
4x  0x  4
2x  2
2
2x  2
+ 1) = 5x  4  2
x 1
Division Algorithm for Polynomials
Let f(x) and d(x) be two polynomials, with
degree of d(x) greater than zero and less
than the degree of f(x). Then there exists
unique polynomials q(x) and r(x) such that
f(x)
= d(x) • q(x) + r(x)
Dividend = Divisor • Quotient + Remainder
where either r(x) = 0 or the degree of r(x) is
less than the degree of d(x).
The polynomial r(x) is called the remainder.
Synthetic Division
• A Quick method of dividing polynomials by
a divisor of the form (x-k).
• Procedure:
1. Arrange the terms of the dividend in
descending order and represent each term
by it’s coefficient only, use zeros to represent
the coefficients of any missing terms.
2. The divisor must be of the form (x-k) or
rewritten in a form equivalent to (x-k).
3. Divide, by multiplying and adding.
4. Synthetic division yields the quotient and the
remainder.
Synthetic Division
2x3+7x2+10x+15 / x+2
Synthetic Division
Use synthetic division to find the quotient:
(x3 – 3x2 + x + 6) ÷ (x – 3)
Step 1: Write the coefficients of the polynomial, and
the k-value of the divisor on the left
3
1
-3
1
6
Synthetic Division
Use synthetic division to find the quotient:
(x3 – 3x2 + x + 6) ÷ (x – 3)
Step 2: Draw a line and write the first coefficient
under the line.
3
1
1
-3
1
6
Synthetic Division
Use synthetic division to find the quotient:
(x3 – 3x2 + x + 6) ÷ (x – 3)
Step 3: Multiply the k-value, 3, by the number
below the line and write the product below the
next coefficient.
3
1
-3
3
1
1
6
Synthetic Division
Use synthetic division to find the quotient:
(x3 – 3x2 + x + 6) ÷ (x – 3)
Step 4: Write the sum of -3 and 3 below the line.
3
1
-3
1
3
0
1
6
Synthetic Division
Use synthetic division to find the quotient:
(x3 – 3x2 + x + 6) ÷ (x – 3)
Repeat steps 3 and 4.
3
1
-3
1
1
3
0
0
1
6
Synthetic Division
Use synthetic division to find the quotient:
(x3 – 3x2 + x + 6) ÷ (x – 3)
Repeat steps 3 and 4.
3
1
-3
1
6
1
3
0
0
1
3
9
Synthetic Division
Use synthetic division to find the quotient:
(x3 – 3x2 + x + 6) ÷ (x – 3)
The remainder is 9 and the resulting numbers are the
coefficients of the quotient.
3
1
-3
1
6
1
3
0
0
1
3
9
x2
9
+1+
x–3
Synthetic Division
(x3 – 3x2 + 4) ÷ (x – 2)
Method 2
Method 1
x2 - x
x–2
–2
x3 – 3x2 + 0x + 4
- (x3 – 2x2)
-x2 + 0x
- (-x2 + 2x)
–2x + 4
- (–2x + 4)
0
2
1
1
-3
0
4
2
-2
-4
-1
-2
0
(x2 – x – 2)
Synthetic Division
• Synthetic division is derived from the recursive nature of
the algorithm. (Recursive – specifies a beginning value
and a way to calculate each successive value)
• This method relies on the relationships between the
coefficients of x in the product of the quotient and divisor.
Example
Use synthetic division to find the quotient and remainder.
 4 x
2 –4
5
 x 4  6 x3  2 x 2  50   ( x  2)
1
6
2
0 50
–8 –14 –16 –28 –56
–4 –7 –8 –14 –28 –6
Note: We must write a
0 for the missing term.
The quotient is – 4x4 – 7x3 – 8x2 – 14x – 28 and the
remainder is –6.
6
4 x 4  7 x3  8 x 2  14 x  28 
x2
Your Turn:
• Use synthetic division to divide
5x3+8x2-x+6 by x+2.
• 5x2-2x+3
• Use synthetic division to divide
x4-10x2-2x+4 by x-3.
11
• x  3x  x  5 
x 3
3
2
Synthetic Division
• Divide 6x3+7x2+x+1 by 2x+3
• 2x+3 = 2(x+3/2)
• (6x3+7x2+x+1)/(2x+3) =
(1/2)[(6x3+7x2+x+1)/(x+3/2)]
5
• 3x  x  2 
2x  3
2
Synthetic Division
REMAINDER THEOREM
Let
4
3
2
f(x) = 2x + x – 3x – 5
What is f(2)?
f(2) = 2(2)4 + (2)3 – 3(2)2 – 5
f(2) = 2(16) + 8 – 3(4) – 5
f(2) = 32 + 8 – 12 – 5
f(2) = 23
This is the same as the remainder when divided by (x – 2):
2
–3
0
–5
2
1
2
4 10 14 28
5 7 14 23
f(2) = 23
The Remainder Theorem
There are two parts of the remainder theorem:
1.
2.
If the polynomial f(x) is divided by (x – a), the remainder will be a
number that is equal to f(a).
1. i.e.. If f(x) is divided by x – 4, f(4) will give the value of the
remainder.
Dividend = (quotient ∙ divisor) + remainder
1. Also can see this as f(x) = [q(x) ∙ (x – a)] + f(a).
2. 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).
– The quotient may also be called a depressed
polynomial because it has one less degree than
the original polynomial.
The Remainder Theorem
That it means:
– Used to find the remainder.
– When dividing a polynomial by a divisor of
the form (x-k), the Remainder Theorem
gives us two ways of calculating the
remainder.
1) Use synthetic division to divide by (x-k).
2) Evaluate the function at f(k).
The Remainder Theorem
Given P(x) = 3x3 – 4x2 + 9x + 5 is divided by x – 6,
find the remainder.
Method 1
6 3
-4
9
Method 2
5
P(6) = 3(6)3 – 4(6)2 + 9(6) + 5
18 84
558
= 3(216) – 4(36) + 54 + 5
3 14 93
563
= 648 – 144 + 54 + 5
= 563
The Remainder Theorem
Use synthetic division and direct substitution to find f(4)
when f(x) = x4 – 6x3 + 8x2 + 5x + 13.
4 1 -6 8 5 13
4 -8 0 20
1 -2 0 5 33
OR
f(4) = 44 – 6(4)3 + 8(4)2 + 5(4) + 13
=256 – 384 + 128 + 20 + 13
f(4) = 33
The Remainder Theorem
Give the factors of x3 – 11x2 + 36x – 36 if one factor is x – 6.
6 1 -11 36 -36
6 -30 36
1 -5 6
0
So, after we divide the polynomial by x – 6
we are left with x2 – 5x + 6 which we can
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 ∙ divisor + remainder.
This would look like f(x) = (x2 – 5x + 6)(x – 6) + 0
or f(x)=(x-2)(x-3)(x-6)..
The Remainder Theorem
• It is usually easiest to evaluate a polynomial for a
specific value of “x” by using synthetic division
and the remainder theorem
• Example: Given
f x   x 5  2 x 4  x 3  5
• Find: f 5
5
f 5  1995
1 2 1 0
0
5
5 15 80 400 2000
1 3 16 80 400 1995
Your Turn:
• Find the remainder when f(x) =4x3+10x2-3x-8 is
divided by (x+1).
• Solution: f(-1)=1, the remainder is 1.
The Remainder Theorem Problems
1.
2.
3.
Use synthetic division to do (4x3 – 9x2 – 10x – 2) divided
by (x – 3). Then write the answer in the form f(x) =
[quotient ∙ divisor] + remainder.
Given f(x) = 4x2 + 6x – 7, find f(-5) by synthetic division
or direct substitution.
Find the factors of x3 + 6x2 – x – 30 if one factors is (x +
5).
Answers:
1) (4x2 + 3x – 1)(x – 3) – 5
2) f(-5) = 63
3) (x + 5), (x – 2), and (x + 3)
The Remainder Theorem Problem
•
•
•
•
Show that x=1 is a solution of the equation 2x3-x2+8=7x+2, and
find any further solutions.
The first part is simple substitution, but you must show the result.
2(1)3-(1)2+8=7(1)+2
9=9, so x=1 is a solution.
Now have to solve the cubic equation 2x3-x2-7x+6=0. We know
x=1 is a solution, so (x-1) is a factor. Using synthetic division,
1
2 -1 -7 6
2 1 -6
2 1 -6 0
So the quadratic factor is 2x2+x-6. This can be factored to
(2x-3)(x+2). Thus the remaining solutions are 3/2 and -2.
Review-The Remainder
Theorem
The Remainder Theorem
If a polynomial f(x) is divided by x – k, the
remainder is equal to f(k).
THE FACTOR THEOREM
k is Zero of a Polynomial Function if P(k) = 0
Example Decide whether the given number is a
zero of P.
(a) 2; P( x)  x 3  4 x 2  9 x  10
3 3
3
2
(b)  2; P( x)  x  x  x
2
2
Analytic Solution
 4 9  10
2 4
10
2 5
0
(a) 2 1
1
(b)  2
3
2
3
2
1
3
4
3
2
0
8  19
19
2  19
The remainder is zero, so
x = 2 is a zero of P.
The remainder is not zero, so
x = –2 is not a zero of P.
The Factor Theorem
•
From the previous example, part (a), we have
P( x)
 x 2  2 x  5  P( x)  ( x  2)( x 2  2 x  5),
x2
indicating that x – 2 is a factor of P(x).
•
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 Factor Theorem
• When a polynomial division results in a
zero remainder
– The divisor is a factor
– f(x) = (x – k) q(x) + 0
• This would mean that f(k) = 0
– That is … k is a zero of the function
• It also means the quotient, q(x), is a
factor of f(x).
Example using the Factor Theorem
Determine whether the second polynomial is a
factor of the first.
P( x)  4 x 3  24 x 2  48x  32; x  2
Solution
Use synthetic division with k = –2.
2 4
4
24
8
16
48
 32
16
32
 32
0
Since the remainder is 0, x + 2 is a factor of P(x),
where
P ( x)  4 x 3  24 x 2  48x  32
 ( x  2)(4 x 2  16 x  16).
Example using the Factor Theorem
Use Remainder Theorem (evaluate f(1)) to determine
whether x – 1 is a factor of x3 – x2 – 5x – 3.
Let x3 – x2 – 5x – 3 = 0
f(1) = (1)3 – (1)2 – 5(1) - 3
f(1) = 1 – 1 – 5 - 3
f(1) = -8
Since f(1) does not equal zero, x – 1 is not a
factor.
Factoring a polynomial using the Factor Theorem
• Factor f(x) = 2x3 + 11x2 + 18x + 9
• Given f(-3)=0
• Since f(-3)=0
• x-(-3) or x+3 is a factor
• So use synthetic division to find the
others!!
Factoring a polynomial using the Factor Theorem
cont.
-3
2
2
11 18 9
-6 -15 -9
5
3 0
So…. 2x3 + 11x2 + 18x + 9 factors to:
(x + 3)(2x2 + 5x + 3)
Now keep factoring gives you:
(x+3)(2x+3)(x+1)
Your Turn:
• Show that (x+3) is a factor of f(x)=x3-19x-30.
Then find the remaining factors f(x).
• Solution: f(-3) = 0, since the remainder is 0,
(x+3) is a factor.
Remaining factors are (x-5) and (x+2).
Example using the Factor Theorem
The function f(x)=x3+ax2-2x+b has (x+1) as a
factor, and leaves a remainder of -3 when
divided by x-2. Find the values of a and b.
-1
1
a -2
b
-1 1-a 1+a
1 a-1 -1-a 1+a+b
1+a+b=0
Solve the system: a+b=1
4a+b=-7
2
1
a
-2
b
2 4+2a 4+4a
1 2+a 2+2a 4+4a+b
4+4a+b=-3
∴ a=-2, b=1
Review-The Factor Theorem
The Factor Theorem
A polynomial f(x) has a factor (x – k), if and
only if f(k)=0.
THE RATIONAL ZERO
THEOREM
The Rational Zeros Theorem
The Rational Zeros Theorem
Let P( x)  an x n  an 1 x n 1    a1 x  a0 , where an  0,
define a polynomial function with integer coefficients.
If p / q is a rational number wri tten in lowest terms,
and if p / q is a zero of P, then p is a factor of the constant
term a0 , and q is a factor of the leading coefficient an .
The Rational Zero Theorem
What it Means:
1. The Rational Zero Theorem allows us to
identify possible rational zeros of the
polynomial. We then check the possible
zeros, using the Remainder and Factor
Theorems, to determine if they are actual
zeros.
Possible rational zeros 
factors of constant term
factors of the leading coefficient
2. Does not agree the polynomial has any
rational zeros.
Example Rational Zeros Theorem
Find all possible rationalzeros: f x  6x 2 10x  4
All factors of - 4 :
All factors of 6 :
 1,  2,  4
 1,  2,  3,  6
p
All possible ratiosof :
q
1 1 1 1 2 2 2 2 4 4 4 4
 , , , , , , , , , , ,
1 2 3 6 1 2 3 6 1 2 3 6
1 1 1
2 1
4 2
 1,  ,  ,  ,  2,  1,  ,  ,  4,  2,  , 
2 3 6
3 3
3 3
1 1 1 2
4
Simplified Possibilit ies :  ,  ,  ,  ,  1,  ,  2,  4
6 3 2 3
3
Your Turn:
• Find all possible rational zeros of
f(x) = 6x3-x2+9x+4.
• Solution: ±1, ±2, ±4, ±1/6, ±1/3,
±1/2, ±2/3, ±4/3
Finding Rational Zeros
• Example: Find the rational zeros for
3
2
f(x) = x + x – 10x + 8
• p/q represents all possible zeros where p are all factors of
the constant, and q are all factors of the leading coefficient.
•
p : all the factors of 8
q : all factors of 1
•
+ 1, + 2, + 4, + 8
+1
• Here are all of the possible zeros for the function:
p
q
+ 1, + 2, + 4, + 8
+1
Possible zeros are
+ 1, + 2, + 4, + 8
So which one do you pick?
Pick any. Find one that is a zero using synthetic division...
3
2
f(x) = x + x – 10x + 8
possible zeros + 1, + 2, + 4, + 8
• Let’s try 1:
1
1
1 –10
8
1
2 –8
1 2 –8
0
1 is a zero of the function
2
The depressed polynomial is x + 2x – 8
2
Find the zeros of x + 2x – 8 by factoring or
(by using the quadratic formula)…
(x + 4)(x – 2) = 0
x = –4, x = 2
The zeros of f(x) are 1, –4, and 2
Finding Zeros of a Polynomial Function
• Use Rational Zeros Theorem to find all possible rational
zeros.
• Use Synthetic Division and Remainder Theorem to try to
find one rational zero (remainder will be zero).
• If “n” is a rational zero, factor original polynomial as
(x – n)q(x).
• Test remaining possible rational zeros in q(x). If one is
found, factor again as in previous step.
• Continue in this way until all rational zeros have been
found.
• See if additional irrational or non-real complex zeros can be
found by solving a quadratic equation.
Example
• Find all zeros of: f x  x  2 x  2 x  3x  2
5
3
2
• Find all solutions to: x  2 x  2 x  3x  2  0
• Rational Zeros Theorem says the only possible
rational zeros are:  1 and  2
• See if -1 is a zero:  1 1 0  2  2  3  2
5
3
2
1 1 1
1
1 1 1 1  2
• Conclusion:
2
0
-1 is a zero,(x  1) is a factorand anotherfactoris :
x
4
- x3 - x 2 - x - 2

Example Continued
• This new factor x - x - x - x - 2 has the same
possible rational zeros:  1 and  2
• Check to see if -1 is also a zero of this:
4
1
• Conclusion:
3
2
1 1 1 1  2
1 2 1
2
1 2 1 2
0
-1 is a zero,(x  1) is a factorand anotherfactoris :
x
3
- 2x2  x - 2

Example Continued
• This new factor x - 2x  x - 2 has as
possible rational zeros:  1 and  2
• Check to see if -1 is also a zero of this:
3
1
• Conclusion:
2
1 2
1
1 3
1 2
3 4
4 6
-1 is NOT a zero,so tryanotherpossible zero: 1
Example Continued
• Check to see if 1 is a zero:
1
1 2 1 2
1 1
0
1 1 0  2
• Conclusion:
-1 is NOT a zero,so tryanotherpossible zero: 2
Example Continued
• Check to see if 2 is a zero:
2
1 2
2
1
0
1 2
0
2
1
0
• Conclusion:
2 is a zero,(x  2) is a factorand anotherfactoris :
x
2

1
Example Continued
• Summary of work done:
f x  x5  2x3  2x 2  3x  2
f x  x 1 x  2x2 1
2
- 1 is a zero of multiplicity two,2 is a zero,and the
other twozeroscan be found by solving:
x 1  0
2
x2 1  0
x 2  1
x  i
Distinct zeros : - 1 (double), 2, i, - i
Your Turn:
•
1.
Find rational zeros of f(x)=x3+2x2-11x-12
List possible
q=1
p=-12
p/q= ±1,± 2, ± 3, ± 4, ± 6, ±12
2.
Test:
1 1 2 -11 -12
p/q=1
1 3 -8
1 3 -8 -20
3.
-1 1 2 -11 -12
p/q=-1
-1 -1 12
1 1 -12 0
Since -1 is a zero: (x+1)(x2+x-12)=f(x)
Factor: (x+1)(x-3)(x+4)=0
x=-1 x=3 x=-4
Your Turn: Find the zeros of g(x) = 6x3 + 4x2 – 14x + 4
Factors of p:
+ 1, + 2, + 4
Factors of q: + 1, + 2, + 3, + 6
Possible zeros are + 1, + 2, + 3, + 4, + 4/3, + 1/2, + 1/3, + 1/6, + 2/3
•
•
4 –14 4
6 10 –4
6 10 –4
0
• 1 is a zero, and 6x2 + 10x – 4 is the depressed
polynomial.
•
Factor 6x2 + 10x – 4 … 2(3x2 + 5x – 2)
We’ll try 1
1
6
2(3x + 1)(x – 2)
Find the zeros… 2(3x + 1)(x – 2) = 0
3x + 1 = 0 x – 2 = 0
x = –1/3
x=2
The zeros are 1, –1/3, and 2
Example :
•
f(x)=10x4-3x3-29x2+5x+12
1.
List: q=10 p=12
p/q= ± 1/1, ± 2/1, ± 3/1, ± 4/1, ± 6/1,±12/1, ± 3/2, ± 1/5, ±
2/5, ± 3/5, ± 6/5, ± 12/5, ± 1/10, ± 3/10, ± 12/10
2.
Check possible zeros using synthetic division:
Check:
-3/2 10 -3 -29 5 12
p/q= -3/2
-15 27 3 -12
10 -18 -2 8
0
Yes it works
* (x+3/2)(10x3-18x2-2x+8)*
(x+3/2)(2)(5x3-9x2-x+4) -factor out GCF
(2x+3)(5x3-9x2-x+4)
-multiply 1st factor by 2
Repeat finding zeros for:
•
g(x)=5x3-9x2-x+4
1.
q=5
p=4
p/q:±1, ±2, ±4, ±1/5, ±2/5, ±4/5
Check possible zeros of depressed equation:
4/5 5 -9 -1 4
p/q=4/5
4 -4 -4
5 -5 -5 0
(2x+3)(x-4/5)(5x2-5x-5)=
(2x+3)(x-4/5)(5)(x2-x-1)= mult.2nd factor by 5
(2x+3)(5x-4)(x2-x-1)=
-now use quad for last*-3/2, 4/5, 1± 5 ,
2
Your Turn:
• Find all zeros of f(x) = x4-6x3-7x2+6x-8
• Solution: 1, -1, 2, 4
Review-Theorems
The Remainder Theorem
If a polynomial f(x) is divided by x – k, the remainder is equal
to f(k).
The Factor Theorem
The polynomial x – k is a factor of the polynomial P(x) if and
only if P(k) = 0.
The Rational Zeros Theorem
Let P( x)  an x n  an 1 x n 1    a1 x  a0 , where an  0,
define a polynomial function with integer coefficients.
If p / q is a rational number wri tten in lowest terms,
and if p / q is a zero of P, then p is a factor of the constant
term a0 , and q is a factor of the leading coefficient an .
Other Tests For Zeros of Polynomials
• Descartes’s Rule of Signs
– Used to help identify the number of real
zeros of a polynomial.
• Upper and Lower Bound Rules
– Used to give an upper or lower bound
of real zeros of a polynomial, which can
help eliminate possible real zeros.
Descartes’ Rule of Signs
Let P(x) be a polynomial function with real coefficients and a
nonzero constant term. The number of positive real zeros of
P(x) is either:
1. The same as the number of variations of sign in P(x), or
2. Less than the number of variations of sign in P(x) by a
positive even integer.
The number of negative real zeros of P(x) is either:
3. The same as the number of variations of sign in P(x), or
4. Less than the number of variations of sign in P(x) by a
positive even integer.
A zero of multiplicity m must be counted m times.
Descartes’ Rule of Signs
• A variation in sign means that two
consecutive (nonzero) coefficients
have opposite signs.
• The polynomial x3-3x+2 has two
variations in sign.
Example: Use Descartes’s Rule of Signs to determine the
possible number of positive and negative real zeros of
f(x) = 2x4 – 17x3 + 35x2 + 9x – 45.
The polynomial has three variations in sign.
+ to –
+ to –
f(x) = 2x4 – 17x3 + 35x2 + 9x – 45
– to +
f(x) has either three positive real zeros or one positive real zero.
f(– x) = 2(– x)4 – 17(– x)3 + 35(– x)2 + 9(– x) – 45
=2x4 + 17x3 + 35x2 – 9x – 45
One change in sign
f(x) has one negative real zero.
f(x) = 2x4 – 17x3 + 35x2 + 9x – 45 = (x + 1)(2x – 3)(x – 3)(x – 5).
Example
• What does Descartes’ rule of signs
tell us about the number of positive
real zeros and the number of negative
real zeros?
P( x)  3x 4  6 x 3  x 2  7 x  2
 3x 4  6 x3  x 2  7 x  2
There are two variations of sign, so
there are either two or zero positive
real zeros to the equation.
Example continued
P( x)  3( x) 4  6( x)3  ( x) 2  7( x)  2
 3x 4  6 x3  x 2  7 x  2
The number of negative real zeros is either
two or zero.
What this tells us;
Total Number of Zeros 4
Positive
2
2
0
0
Negative 2
0
2
0
Nonreal
0
2
2
4
Your Turn:
Example Determine the possible number of positive real zeros and
negative real zeros of P(x) = x4 – 6x3 + 8x2 + 2x – 1.
We first consider the possible number of positive zeros by observing
that P(x) has three variations in signs.
+ x4 – 6x3 + 8x2 + 2x – 1
1
2
3
Thus, by Descartes’ rule of signs, f has either 3 or 3 – 2 = 1 positive
real zeros.
For negative zeros, consider the variations in signs for P(x).
P(x) = (x)4 – 6(x)3 + 8(x)2 + 2(x)  1
= x4 + 6x3 + 8x2 – 2x – 1
Since there is only one variation in sign,
P(x) has only one negative real root.
Total number of zeros 4
Positive: 3 1
Negative: 1 1
Nonreal: 0 2
Upper and Lower Bound Rules
Let P(x) define a polynomial function of degree n  1 with
real coefficients and with a positive leading coefficient. If
P(x) is divided synthetically by x – c, and
(a)
if c > 0 and all numbers in the bottom row of the synthetic
division are nonnegative, then P(x) has no zero greater
than c;
(b)
if c < 0 and the numbers in the bottom row of the synthetic
division alternate in sign (with 0 considered positive or
negative, as needed), then P(x) has no zero less than c.
Example:
Show that the real zeros of P(x) = 2x4 – 5x3 + 3x + 1
satisfy the following conditions.
(a) No real zero is greater than 3.
(b) No real zero is less than –1.
Solution
a) c > 0
3 2 5 0 3 1
6 3 9 36
2 1 3 12 37
b) c < 0
1 2  5 0 3 1
6 7 7 4
2 7 7 4 5
All are nonegative.
No real zero greater than 3.
The numbers alternate in sign.
No zero less than 1.
How to Use the Upper and Lower Bound
Rules
• Finding the real zeros of
f x  x3  2x 2  3x  6
we check the possible zero 2 by synthetic
division:
2
1
1
2 3 6
2
8 10
4
5
4
• Notice bottom row. What does this tell us
about zeros?
f  x  has no zero greater than 2
• Now we don’t need to check any possible
zeros greater than 2.
How to Use the Upper and Lower Bound
Rules
• Using the same function:
f x  x3  2x 2  3x  6
check the possible zero -3
3
2 3 6
3
3
0
1 1 0  6
1
• Notice bottom row. What does this tell us about
zeros? f x has no zero less than 3
• So we don’t need to check any possible zeros
less than -3.