section 2.3 polynomial and synthetic division
DESCRIPTION
Section 2.3 Polynomial and Synthetic Division. What you should learn. How to use long division to divide polynomials by other polynomials How to use synthetic division to divide polynomials by binomials of the form ( x – k ) How to use the Remainder Theorem and the Factor Theorem. - PowerPoint PPT PresentationTRANSCRIPT
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Section 2.3 Polynomial and Synthetic Division
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What you should learn
• How to use long division to divide polynomials by other polynomials
• How to use synthetic division to divide polynomials by binomials of the form
(x – k)• How to use the Remainder Theorem and the
Factor Theorem
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641 23 xxxx
2x1. x goes into x3? x2 times.2. Multiply (x-1) by x2.
23 xx 220 x x4
4. Bring down 4x.
5. x goes into 2x2? 2x times.
x2
6. Multiply (x-1) by 2x.
xx 22 2 x60
8. Bring down -6.
69. x goes into 6x?
6
66 x0
3. Change sign, Add.
7. Change sign, Add
6 times.
11. Change sign, Add .10. Multiply (x-1) by 6.
3 2x x
22 2x x
6 6x
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Long Division.
1583 2 xxxx
xx 32
155 x
5
155 x0
)5)(3( xx
Check
15352 xxx
1582 xx
2 3x x
5 15x
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Divide.
3 273
xx
33 27x x
3 23 0 0 27x x x x
2x
3 23x x3 23x x 23 0x x
3x
23 9x x23 9x x 9 27x
9
9 27x 9 27x 0
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Long Division.
824 2 xxxx
xx 42
82 x
2
82 x0
)4)(2( xx
Check
8242 xxx
822 xx
2 4x x
2 8x
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Example
2026 2 ppp
p
pp 62
204 p
4
244 p44
6
44)6()4)(6(p
ppp
Check
4424642 ppp
2022 pp
62022
ppp
644
p
2 6p p
4 24p
=
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2022 pp
62022
ppp
6444
p
p
)6(6
4464
pp
pp
4464 pp2022 pp
)()()(
)()(
xdxrxq
xdxf
)()()()( xrxqxdxf
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The Division Algorithm
If f(x) and d(x) are polynomials such that d(x) ≠ 0, and the degree of d(x) is less than or equal to the degree of f(x), there exists a unique polynomials q(x) and r(x) such that
Where r(x) = 0 or the degree of r(x) is less than the degree of d(x).
)()()()( xrxqxdxf
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Proper and Improper
• Since the degree of f(x) is more than or equal to d(x), the rational expression f(x)/d(x) is improper.
• Since the degree of r(x) is less than than d(x), the rational expression r(x)/d(x) is proper.
)()()(
)()(
xdxrxq
xdxf
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Synthetic DivisionDivide x4 – 10x2 – 2x + 4 by x + 3
1 0 -10 -2 4-3
1
-3
-3
+9
-1
3
1
-3
1
3
4210 24
xxxx
31
x
13 23 xxx
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Long Division.
823 2 xxxx
xx 32
8x
1
3x582)( 2 xxxf
xx 32
3 x
)3(f 8)3(2)3( 2 869
5
1 -2 -83
1
3
1
3
-5
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The Remainder Theorem
If a polynomial f(x) is divided by x – k, the remainder is r = f(k).
82)( 2 xxxf)3(f 8)3(2)3( 2
869 5
823 2 xxxx
xx 32
8x
1
3x5
xx 32
3 x
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The Factor TheoremA polynomial f(x) has a factor (x – k) if and only
if f(k) = 0.Show that (x – 2) and (x + 3) are factors of
f(x) = 2x4 + 7x3 – 4x2 – 27x – 18
2 7 -4 -27 -18+2
2
4
11
22
18
36
9
18
0
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Example 6 continued
Show that (x – 2) and (x + 3) are factors of f(x) = 2x4 + 7x3 – 4x2 – 27x – 18
2 7 -4 -27 -18+2
2
4
11
22
18
36
9
18
-3
2
-6
5
-15
3
-9
0 1827472 234 xxxx)2)(918112( 23 xxxx)3)(2)(352( 2 xxxx)3)(2)(1)(32( xxxx
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Uses of the Remainder in Synthetic Division
The remainder r, obtained in synthetic division of f(x) by (x – k), provides the following information.
1. r = f(k)2. If r = 0 then (x – k) is a factor of f(x).3. If r = 0 then (k, 0) is an x intercept of the
graph of f.
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Fun with SYN and the TI-83
• Use SYN program to calculate f(-3)• [STAT] > Edit• Enter 1, 8, 15 into L1, then [2nd][QUIT]• Run SYN• Enter -3
158)( 2 xxxf )3(f
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Fun with SYN and the TI-83
• Use SYN program to calculate f(-2/3)• [STAT] > Edit• Enter 15, 10, -6, 0, 14 into L1, then [2nd]
[QUIT]• Run SYN• Enter 2/3
1461015)( 234 xxxxf
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2.3 Homework
• 1-67 odd