Download 4.1 Polynomial Functions

4.1 Polynomial Functions
OBJECTIVES:
1.
2.
3.
4.
DEFINE A POLYNOMIAL.
DIVIDE POLYNOMIALS.
APPLY THE REMAINDER AND FACTOR THEOREMS,
AND THE CONNECTIONS BETWEEN REMAINDERS
AND FACTORS.
DETERMINE THE MAXIMUM NUMBER OF ZEROS OF
A POLYNOMIAL.
Polynomials
Characteristics of Polynomials:
 A sum or difference of monomials
 All exponents are whole numbers
 No variable contained in the denominator
 No variable is under a radical
Polynomials
1
2
x 2  5
7
2y  7y  1
x 3  6x 2 
y
15
y
10
w  6.7
12
Not Polynomials
1
3
m 2 m
5
p2   3
p
Degree of a Polynomial
Example #1
Polynomial Division
 Divide:
2x  25x  30x  5 by x  4
4
2
Expand the dividend and fill in missing terms, then
follow the same steps as when dividing real numbers:
divide, multiply, subtract, bring down.
2x3
x  4 2 x 4  0 x 3  25 x 2  30 x  5
(2x4 − 8x3)
Bring down
Subtract
8x3 – 25x2
Divide
2x4
 2x3
x
Multiply
2 x3  x  4 
 2 x 4  8 x3
Once this cycle is made once, it is repeated until the degree of the
remainder is less than the degree of the dividend.
Example #1
Polynomial Division
 Divide:
2x  25x  30x  5 by x  4
4
2
2x3 + 8x2
x  4 2 x 4  0 x 3  25 x 2  30 x  5
(2x4 − 8x3)
–
(8x3 – 32x2)
8x3
Subtract
25x2
7x2
Bring
down
– 30x
Divide
8 x3
 8x2
x
Multiply
8 x 2 x  4
 8 x3  32 x 2
Example #1
Polynomial Division
 Divide:
2x  25x  30x  5 by x  4
4
2
2x3 + 8x2 + 7x
x  4 2 x 4  0 x 3  25 x 2  30 x  5
(2x4 − 8x3)
Bring
–
(8x3 – 32x2)
8x3
Subtract
25x2
7x2
(7x2
down
– 30x
– 28x)
−2x + 5
Divide
7 x2
 7x
x
Multiply
7 x x  4 
 7 x 2  28 x
Example #1
Polynomial Division
 Divide:
2x  25x  30x  5 by x  4
4
2
2x3 + 8x2 + 7x − 2
x  4 2 x 4  0 x 3  25 x 2  30 x  5
(2x4 − 8x3)
 2x
 2
x
8x3 – 25x2
(8x3 – 32x2)
The degree of the
remainder is less
than that of the
divisor so the cycle
is completed.
7x2
(7x2
Subtract
Divide
– 30x
– 28x)
−2x + 5
(−2x + 8)
−3
Multiply
 2 x  4 
 2 x  8
Remainder
Example #1
Polynomial Division
 Divide:
2x  25x  30x  35 by x  4
4
2

2x3 + 8x2 + 7x − 2 x  4
x  4 2 x 4  0 x 3  25 x 2  30 x  5
(2x4 − 8x3)
Finally the remainder is
put over the divisor and
added to the quotient.
8x3 – 25x2
(8x3 – 32x2)
7x2 – 30x
(7x2 – 28x)
−2x + 5
(−2x + 8)
−3
Example #2
Synthetic Division
 Divide using synthetic division:
x  5x  2x  6x  17 by x  5
4
3
2
Caution! Synthetic division only works when the divisor
is a first-degree polynomial of the form (x − a).
Set the divisor
equal to 0 and
solve. Place it in
the left corner in
a half-box
x50
x  5
−5
1
5
−5
1
0
2
6
−17
Bring down the first digit. Add the
column and
Multiply it by the -5.
repeat the
Write the answer
process.
under the next digit.
Write the leading
coefficients of the
dividend in order of
degree next to the
half-box. Include 0’s
for missing terms if
necessary.
Example #2
Synthetic Division
 Divide using synthetic division:
x  5x  2x  6x  17 by x  5
4
3
−5
Multiply the sum
again by the -5.
2
1
5
−5
2
0
1
0
2
6
−17
Write the answer
under the next digit.
Add the column and
repeat the process.
Example #2
Synthetic Division
 Divide using synthetic division:
x  5x  2x  6x  17 by x  5
4
3
−5
Multiply the sum
again by the -5.
2
1
5
−5
1
0
2 6 −17
0 −10
2 −4
Write the answer
under the next digit.
Add the column and
repeat the process.
Example #2
Synthetic Division
 Divide using synthetic division:
x  5x  2x  6x  17 by x  5
4
3
−5
Multiply the sum
again by the -5.
2
1
5
−5
1
0
2 6 −17
0 −10 20
2 −4
3
Write the answer
under the next digit.
Add the column.
The last digit forms
the remainder.
Example #2
Synthetic Division
 Divide using synthetic division:
x  5x  2x  6x  17 by x  5
4
3
2
After using up all terms, the quotient needs the variables added back into
the final answer. The first term is always one degree less than the dividend.
Remember to place the remainder over the original divisor.
−5
1
5
−5
1
0
2 6 −17
0 −10 20
2 −4
 1x 3  0 x 2  2 x  4 
 x3  2x  4 
3
x5
3
3
x5
Example #3
Factors Determined by Division
Determine if 3x  2 is a factor of 3x  15x  2x  10.
2
x
3x 2  0 x  2 3x 3  15 x 2  2 x  10
Bring
(3x3 + 0x2 – 2x)
down
Subtract
2
15x + 0x −10
3
2
Divide
3
3x
3x
2
x
Multiply

x 3x 2  0 x  2

 3x 3  0 x 2  2 x
Example #3
Factors Determined by Division
Determine if 3x  2 is a factor of 3x  15x  2x  10.
2
3x 2  0 x  2 3x 3  15 x 2
(3x3 + 0x2
15x2
(15x2
Subtract
3
x+5
 2 x  10
– 2x)
+ 0x −10
+ 0x – 10)
0
2
Divide
2
15 x
5
2
3x
Multiply

5 3x 2  0 x  2

 15 x 2  0 x  10
Example #3
Factors Determined by Division
Determine if 3x  2 is a factor of 3x  15x  2x  10.
2
3x 2  0 x  2 3x 3  15 x 2
(3x3 + 0x2
15x2
(15x2
3
x+5
 2 x  10
– 2x)
+ 0x −10
+ 0x – 10)
0
Because the remainder is 0, the divisor divides evenly into the
dividend and is then said to be a factor of the dividend.
2
Remainder Theorem
 If a polynomial f(x) is divided by x – c, then the
remainder is f(c).
Example #4
The remainder when dividing by x – c:
Find the remainder when x 87  2x 36  8 is divided by x  1.
Using the remainder theorem, this problem can be performed
without actually dividing the dividend by the divisor.
First find the value of
c in the divisor x – c.
x 1 0
x 1
Next evaluate f(c).
87
36
f 1  1  21  8
1 2  8
7
The remainder is therefore 7.
Example #5
The remainder when dividing by x + k
Find the remainder when 2x4 + 4x3 + 5x – 6 is divided by x + 3.
x 30
x  3
f  3  2 34  4 33  5 3  6
 281  4 27   15  6
 162  108  15  6
 33
Factor Theorem
 A polynomial function f(x) has a linear factor
x – a if and only if f(a) = 0.
The idea in this theorem was actually introduced in Example #3.
The difference is that this theorem only works for linear factors,
in which case division is not necessary, otherwise you must use
division to check if they are factors.
Example #6
The Factor Theorem
Show that x  2 is a factor of x 3  5x 2  2x  24 by using the Factor Theorem.
Find q(x) such that (x  2)q(x)  x 3  5x 2  2x  24.
f 2  23  522  22  24
x20
 8  20  4  24
x2
0
Since f (2) = 0, then it is a factor of the polynomial.
Next we’ll find the actual quotient using synthetic division.
2
1
1
5
2
−2
14
−24
24
7
12
0
x  7 x  12
2
The answer can be checked
by multiplication.
x  2x 2  7 x  12 
 x 3  7 x 2  12 x  2 x 2  14 x  24
 x 3  5 x 2  2 x  24
Fundamental Polynomial Connections
If one of the following are true, all are true:
A. r is a zero of the function f
B. r is an x-intercept of the graph of the function f
C. x = r is a solution, or root, of the function f(x) = 0
D. x – r is a factor of the polynomial f(x)
Example #7
Fundamental Polynomial Connections
For f(x)  2x 3  7x 2  20x  25, find the following:
5
0
2
5

2 x    20 
2

2x  5  0
x
a) the x-intercepts of the graph of f
**Note**:
b) the zeros of f
x  2 .5  0
c) the solutions to 2x 3  7x 2  20x  25  0
d) the linear factors with real coefficients of 2x 3  7x 2  20x  25
y
From the graph it is clear that the x-intercepts
are at −2.5, 1, & 5.
50
40
30
The zeros of f are also −2.5, 1, & 5.
20
10
–5
–4
–3
–2
–1
–10
1
2
3
4
5
x
The solutions are x = −2.5, x = 1, & x = 5.
–20
–30
–40
–50
Setting each solution equal to 0 we obtain the
linear factors of (2x + 5), (x – 1), & (x – 5).
Example #8
A Polynomial with Specific Zeros
Find three polynomials of different degrees that have
-2, 1, and 3 as zeros.
degree of 3  g  x    x  2 x  1 x  3
degree of 4  h x    x  22  x  1 x  3
degree of 5  k  x    x  22  x  12  x  3
The simplest polynomial must be of degree 3 to have all the zeros
requested. For a degree of 4, one (not necessarily the factor
shown) of the factors must repeat twice, but highest exponent (if it
were multiplied out) would still only be 4. Finally, for a degree of 5,
two of the factors must repeat.
Example #8
A Polynomial with Specific Zeros
Find three polynomials of different degrees that have
-2, 1, and 3 as zeros.
degree of 5  k x   x  22 x  12 x  3


degree of 5  px   7x  2x  1x  3 x 2  1
Here, an alternative polynomial of degree 5 is shown that retains
the same zeros as the one above it. First of all, the 7 out front is a
constant and has NO effect on where the graph crosses the x-axis.
This could be any real number, even a negative.
Secondly, the final group of (x2 + 1) has imaginary roots at ±i.
When everything is multiplied out, the resulting polynomial will
still have a degree of 5, but the imaginary roots will have no effect
on where the graph crosses the x-axis either.
Number of Zeros
A polynomial of degree n has
at most n distinct real zeros.
Example #8
Determining the Number of Zeros
 Find the maximum number of zeros that the
polynomial could contain.
f ( x)  6 x  2 x  x  7
5
3
The degree of this polynomial is 5.
At most, this polynomial can have only 5 distinct real zeros.
This means that at most, this polynomial will cross the x-axis 5 times.