OBJECTIVES: 1. DEFINE A POLYNOMIAL. 2. DIVIDE POLYNOMIALS. 3. APPLY THE REMAINDER AND FACTOR...
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Transcript of OBJECTIVES: 1. DEFINE A POLYNOMIAL. 2. DIVIDE POLYNOMIALS. 3. APPLY THE REMAINDER AND FACTOR...
O B J E C T I V E S :
1. D E F I N E A P O LY N O M I A L .2. D I V I D E P O LY N O M I A L S .3. A P P LY T H E R E M A I N D E R A N D FA C T O R T H E O R E M S ,
A N D T H E C O N N E C T I O N S B E T W E E N R E M A I N D E R S A N D FA C T O R S .
4. D E T E R M I N E T H E M A X I M U M N U M B E R O F Z E R O S O F A P O LY N O M I A L .
4.1 Polynomial Functions
Polynomials
Characteristics of Polynomials:A sum or difference of monomialsAll exponents are whole numbersNo variable contained in the denominatorNo variable is under a radical
Polynomials Not Polynomials
x3 6x2 12
y 15 y10 7
w 6.7
12
x 2 5
2y
1
3 7y 1
m 2 m
p 2 5p
3
Example #1Polynomial Division
Divide: 2x 4 25x 2 30x 5 by x 4Expand the dividend and fill in missing terms, then follow the same steps as when dividing real numbers: divide, multiply, subtract, bring down.
53025024 234 xxxxx2x3
8x3
34
22
xx
x
Divide
34
3
82
42
xx
xx
Multiply
(2x4 − 8x3)
Subtract
Bring down
– 25x2
Once this cycle is made once, it is repeated until the degree of the remainder is less than the degree of the dividend.
Example #1Polynomial Division
Divide: 2x 4 25x 2 30x 5 by x 4
53025024 234 xxxxx2x3
8x32
3
88
xx
x
Divide
23
2
328
48
xx
xx
Multiply
(2x4 − 8x3)
Subtract
Bringdown– 25x2
+ 8x2
(8x3 – 32x2)
7x2 – 30x
Example #1Polynomial Division
Divide: 2x 4 25x 2 30x 5 by x 4
53025024 234 xxxxx2x3
8x3x
x
x7
7 2
Divide
xx
xx
287
472
Multiply
(2x4 − 8x3)
Subtract
Bringdown
– 25x2
+ 8x2
(8x3 – 32x2)
7x2 – 30x
+ 7x
(7x2 – 28x)
−2x + 5
Example #1Polynomial Division
Divide: 2x 4 25x 2 30x 5 by x 4
53025024 234 xxxxx2x3
8x32
2
x
x
Divide
82
42
x
x
Multiply
(2x4 − 8x3)
Subtract
– 25x2
+ 8x2
(8x3 – 32x2)
7x2 – 30x
+ 7x
(7x2 – 28x)
−2x + 5
− 2
(−2x + 8)
−3 Remainder
The degree of the remainder is less than that of the divisor so the cycle is completed.
Example #1Polynomial Division
Divide: 2x 4 25x 2 30x 5 by x 4
53025024 234 xxxxx2x3
8x3
4
3
x
(2x4 − 8x3)
– 25x2
+ 8x2
(8x3 – 32x2)
7x2 – 30x
+ 7x
(7x2 – 28x)
−2x + 5
− 2
(−2x + 8)
−3
Finally the remainder is put over the divisor and added to the quotient.
Example #2Synthetic Division
Divide using synthetic division:
x 4 5x 3 2x 2 6x 17 by x 5Caution! Synthetic division only works when the divisor is a first-degree polynomial of the form (x − a).
5
05
x
x
−5 1 5 2 6 −17Set the divisor equal to 0 and solve. Place it in the left corner in a half-box
Write the leading coefficients of the dividend in order of degree next to the half-box. Include 0’s for missing terms if necessary.
1Bring down the first digit.
−5
Add the column and repeat the process.
0
Multiply it by the -5.Write the answer under the next digit.
Example #2Synthetic Division
Divide using synthetic division:
x 4 5x 3 2x 2 6x 17 by x 5
−5 1 5 2 6 −17
1
−5
Add the column and repeat the process.
0
Multiply the sum again by the -5.
Write the answer under the next digit.
0
2
Example #2Synthetic Division
Divide using synthetic division:
x 4 5x 3 2x 2 6x 17 by x 5
−5 1 5 2 6 −17
1
−5
Add the column and repeat the process.
0
Multiply the sum again by the -5.
Write the answer under the next digit.
0
2 −4
−10
Example #2Synthetic Division
Divide using synthetic division:
x 4 5x 3 2x 2 6x 17 by x 5
−5 1 5 2 6 −17
1
−5
Add the column. The last digit forms the remainder.
0
Multiply the sum again by the -5.
Write the answer under the next digit.
0
2 −4
−10
3
20
Example #2Synthetic Division
Divide using synthetic division:
x 4 5x 3 2x 2 6x 17 by x 5
−5 1 5 2 6 −17
1
−5
0
0
2 −4
−10
3
20
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
342
5
34201
3
23
xxx
xxxx
Example #3Factors Determined by Division
Determine if 3x 2 2 is a factor of 3x 3 15x 2 2x 10.
102153203 232 xxxxx
xx
x
2
3
3
3
xxx
xxx
203
20323
2
x
(3x3 + 0x2 – 2x)Divide
Multiply
Subtract 15x2 + 0x −10
Bring down
Example #3Factors Determined by Division
Determine if 3x 2 2 is a factor of 3x 3 15x 2 2x 10.
102153203 232 xxxxx
53
152
2
x
x
10015
20352
2
xx
xx
x
(3x3 + 0x2 – 2x)Divide
MultiplySubtract
15x2 + 0x −10
+ 5
(15x2 + 0x – 10)
0
Example #3Factors Determined by Division
Determine if 3x 2 2 is a factor of 3x 3 15x 2 2x 10.
102153203 232 xxxxxx
(3x3 + 0x2 – 2x)
15x2 + 0x −10
+ 5
(15x2 + 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.
Example #4The 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.
1
01
x
x
7
821
81211 3687
f
First find the value of c in the divisor x – c.
Next evaluate f(c).
The remainder is therefore 7.
Example #5The remainder when dividing by x + k
3
03
x
x
33
615108162
615274812
63534323 34
f
Find the remainder when 2x4 + 4x3 + 5x – 6 is divided by x + 3.
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 #6The 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.
2
02
x
x
0
244208
24222522 23
f
Since f (2) = 0, then it is a factor of the polynomial. Next we’ll find the actual quotient using synthetic division.
2 1 5 −2 −24
1 7 12 0
2 14 24
1272 xx
The answer can be checked by multiplication.
2425
24142127
1272
23
223
2
xxx
xxxxx
xxx
Fundamental Polynomial Connections
If one of the following are true, all are true:
A. r is a zero of the function fB. r is an x-intercept of the graph of the
function fC. x = r is a solution, or root, of the function
f(x) = 0D. x – r is a factor of the polynomial f(x)
Example #7Fundamental Polynomial Connections
For f(x) 2x 3 7x 2 20x 25, find the following:
a) the x-intercepts of the graph of f
b) the zeros of f
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
1 2 3 4 5–1–2–3–4–5 x
10
20
30
40
50
–10
–20
–30
–40
–50
y
From the graph it is clear that the x-intercepts are at −2.5, 1, & 5.
The zeros of f are also −2.5, 1, & 5.
The solutions are x = −2.5, x = 1, & x = 5.Setting each solution equal to 0 we obtain the linear factors of (2x + 5), (x – 1), & (x – 5).
052
022
52
02
5
05.2
x
x
x
x
**Note**:
Example #8A Polynomial with Specific Zeros
Find three polynomials of different degrees that have
-2, 1, and 3 as zeros. 3125ofdegree
3124ofdegree
3123ofdegree
22
2
xxxxk
xxxxh
xxxxg
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 #8A Polynomial with Specific Zeros
Find three polynomials of different degrees that have
-2, 1, and 3 as zeros. 131275ofdegree
3125ofdegree2
22
xxxxxp
xxxxk
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.