Chapter 13
Roots of Polynomials
1
Contents
• Remainder and Factor Theorem
• Fundamental Theorem of Algebra
• Rational and Irrational Roots
• Conjugate Roots
• Partial Fractions
2
Division of Polynomials
• This is not officially on the schedule, but you will need to know
it for future sections.
• There are two choices: long division and synthetic division
3
Long Division
Integers
672 / 21
3
21 ) 672
63
42
32
21 ) 672
63
42
42
0
Polynomials
(6x2+ 7x + 2) / (2x + 1)
3x
(2x + 1) ) 6x2+ 7x + 2
6x2+ 3x
4x + 2
3x + 2
(2x + 1) ) 6x2+ 7x + 2
6x2+ 3x
4x + 2
4x + 2
0
Example
• (7y + 6 y3 + 2 – 19y2) / (3y – 2) =
• First reorder:
(6 y3– 19y2 + 7y + 2 ) / (3y – 2)
= 2y2 – 5y - 1
What About Remainders????
• (x2 + 2x – 13) / (x + 5) =
x-3
(x + 5) ) x + 2x- 13
x + 5x
-3x – 13
-3x – 15
2
Write x – 3 + 2 / (x + 5), just like we do with numbers
Synthetic Division
x3 + 2x2 – 4 / (x-3)
Rewrite 3| 1 2 0 -4, using just the coefficients;
note the sign change on the 3!
Bring down the 1:
3| 1 2 0 -4
3 15 45
1 5 15 41
since 3x1 = 3, 2+3 = 5
then 3x5=15 and 0+15 = 15
then 3x15 = 45 and -4 + 45 = 41
Quotient is x2 + 5x + 15, remainder 41
7
Example
Divide x3 – 6x + 4 by x-2
8
Solution
Divide x3 – 6x + 4 by x-2
2|
1
1
0 -6 4
2
4 -4
2
-2
0
(x-2)(x2+2x -2)
9
What about a remainder?
Divide x4 -2x3 + 5x2 -4x + 3 by (x+1)
-1 | 1 -2 5 -4 3
-1 3 -8 12
1 -3 8 -12 15
The remainder is 15
10
Chapter 13.2
Remainder and Factor Theorem
11
Basics
• Polynomial equation: anxn + an-1xn-1 + …a1x + a0 = 0
• Solution or root is a number r that makes the equation true
• r is also called a root
12
Multiplicity of a Root
• Consider x3 - 2x2 +x = 0
• Factor to get x(x-1)(x-1)
Roots are x=0, x=1
• x=1 is called a repeated, or double root;
it appears in two factors
• We call a root that appears in k factors a root of multiplicity k
13
Example
• (x-4)3(x-1)4(x+5)7 = 0
 4 is a root of multiplicity 3
 1 is a root of multiplicity 4
 -5 is a root of multiplicity 7
• If we have a repeated root, then the graph of the polynomial is
tangent to the x axis at the root
• If the multiplicity is odd, it does cross the axis
• If the multiplicity is even, it does not cross, just touches
14
f(x) = (x+1)(x-2)2
10
5
0
-6
-4
-2
0
2
4
6
-5
-10
15
f(x) = (x+1) 3(x-2)
Even for an odd multiplicity, the curve is tangent at the root
16
Remainder Theorem
• If f(x) is divided by (x-r) then the remainder is f(r)
• Proof:
f(x) = (x-r) q(x) + r(x), where r is the remainder
Since r(x) is the remainder, it has degree less than (x-r),
so has degree zero; r(x) must be a constant
f(x) = (x-r)q(x) + c
if x = r, we have f(r) = (r-r)q(r) + c = c,
c is the remainder, c = f(r)
17
Example
• Show that (x-a) is a factor of f(x) = xn - an
18
Solution
• Show that (x-a) is a factor of f(x) = xn - an
• When f(x) is divided by (x-a), the remainder is f(a)
• f(a) = an-an = 0
• Since the remainder is zero, x-a is a factor
19
Factor Theorem
• If f(x) is a polynomial, and f(r) = 0, then x-r is a factor of f(x)
20
Using the Factor Theorem
• Solve x4 + 2x3 – 7x2 – 20x – 12 given that x=3 and x=-2
are roots
21
Solution
First step, x=3 is a factor
3| 1 2 -7 -20 -12
3 15 24 12
1 5 8
4
0
Now for x = -2
-2| 1 5 8 4
-2 -6 -4
1 3
2 0
Leaves x2 + 3x + 2 = 0 = (x+2)(x+1)
22
Recap
• Polynomial degree n: f(x) = anxn + an-1xn-1 + …a1x + a0 , an ≠0
• Root, xo satisfies f(xo) = 0
• Multiplicity of a root; how often the factor (x-xo) occurs
f(x) is tangent to the axis at a multiple root
• Remainder theorem: if f(x) is divided by (x-r), the remainder
is f(r)
• If f(r) = 0, then (x-r) is a factor of f(x)
23
Different Approach
• Find f(x) such that -1, 4, and 5 are roots
24
Solution
• Find f(x) such that -1, 4, and 5 are roots
• f(x) = (x+1)(x-4)(x-5)
25
Example
• Find a polynomial such that 3 is a root and -4 is a root of
multiplicity 2.
26
Solution
• Find a polynomial such that 3 is a root and -4 is a root of
multiplicity 2.
• f(x) = (x-3)(x+4)2
27
Section 13.3
Fundamental Theorem of Algebra
28
The Theorem
Every polynomial of the form anxn + an-1xn-1 + …a1x + a0 = 0, n≥1
and an≠ 0 has at least one root in the complex number system;
that root may be a real number.
Additionally, a polynomial of degree n has n roots (where a root
of multiplicity k is counted as k roots).
We will assume the coefficients are real, but the theorem holds if
they are complex. The powers must be integers in order to have
a polynomial
Note: No statement is made about our ability to find the root!
29
Relating Roots and Coefficients
• If r1 and r2 are roots of x2 + bx + c = 0, show r1r2 = c, and
r1 + r2 =- b
30
Solution
• If r1 and r2 are roots of x2 + bx + c = 0, show r1r2 = c, and
r1 + r2 = -b
Since are roots, (x-r1)(x-r2) = 0
(x-r1)(x-r2) = x2 – (r1+r2) + r1r2 = 0,
clearly c = r1r2 and –b = r1 + r2
31
If we have more roots?
• x3 + bx2 + cx + d = 0
• We still have:
r1 + r2 + r3 = -b,
r1r2+r2r3+r1r3=c, and
r1r2 r3 =-d
• We can use this to find a root or to check it
32
Examples
• Given the roots and the multiplicity, find f(x)
Root
Multiplicity
0
2
2
1
-3
2
33
Solution
• Given the roots and the multiplicity, find f(x)
Root
Multiplicity
0
2
2
1
-3
2
• f(x) = x2 (x-2)(x+3)2 = x2 (x-2)(x2+6x+9) = x2 (x3+4x2-3x -18)
(x5 + 4x4 – 3x3 – 18x2)
• Note: In the portion x3+bx2-cx –d.
18 = 2 (-3)(-3) = d (product of roots)
-3 = (2(-3) + 2(-3) + (-3)(-3) ) =(-12 + 9) = -3 = c
(sum of pair products)
-b = 4 = -(2 + -3 + -3) = 4 (-sum of roots)
34
Chapter 13.4
Rational and Irrational Roots
35
Rational Roots Theorem
• For the polynomial anxn + an-1xn-1 + …a1x + a0 = 0,
n≥1 and an≠ 0
If p/q is a rational root of the equation, and p and q have no
common factors other than ± 1 (relatively prime)
Then p is a factor of a0 and q is a factor of an.
36
Applying the Theorem
• Find the roots: 2x3 – x2 – 9x -4= 0
37
Solution
• Find the roots: 2x3 – x2 – 9x -4= 0
• If p and q are roots, p is a factor of -4 and q of 2
• p = ±1, ±2, ±4
• q = ±1, ±2
• Possible roots are p/q: ±1, ±1/2, ±2, ±4
• Using synthetic division, -1/2 is a root, leaving 2x2 – 2x – 8=0
• The solution to this is (1 ± 17 )/2
• We know that the equation has 3 roots, so there are no others
38
Realistically?
• There is no guarantee that an equation has a rational root, so
this method may be useless
• For example: x3 + 8x2 – 1 = 0
look for p and q, p divides -1, q divides 1
Only possible solutions are ± 1, neither of which works
• This equation has no rational roots
39
Examples
• What are the possible rational roots of x4 – x3 + 10x2 – 24 =0
40
Solution
• 24 has roots ± 1, 2, 3, 4, 6, 8, 12
41
Example
• What are the possible rational roots of 18x4 – 10x3 + x2 – 4 =0
•
42
Solution
• What are the possible rational roots of 18x4 – 10x3 + x2 – 4 =0
• Factors of 18 are ± 1, 2, 3, 6, 9, 18
• Factors of 4 are ± 1, 2, 4
• Possible roots are ± 1, 2, 4, ½, 1/3, 2/3, 4/3, 1/6, 1/9, 2/9, 4/9,
1/18
43
Example
• Show there are no rational roots to x3 -3x + 1 - 0
44
Solution
• Show there are no rational roots to x3 -3x + 1 = 0
• Possible roots are 1, -1; 1 -3+ 1≠0, -1 + 3 - 1 ≠ 0
45
Upper and Lower Bounds
• For the polynomial anxn + an-1xn-1 + …a1x + a0 = 0,
n≥1 and an> 0 and all of the coefficients are real
– Divide f(x) by x-b, b>0. If there are no negative
coefficients, b is an upper bound for the roots of
the polynomial
– Divide f(x) by x-b, b<0, and the coefficients are alternating
positive and negative, then b is a lower bound for real
roots
46
Example
• Find the rational roots for:
¼ x4 – ¾ x3 + 17
4
x2 + 4x + 5 = 0
47
Solution
• Find the rational roots for:
¼ x4 – ¾ x3 + 17
4
x2 + 4x + 5 = 0
• First multiply by 4:
x4 – 3 x3 + 17x2 + 16x + 20 = 0
• Factors of 20 are ± 1, 2, 4, 5, 10, 20, so are the possible roots
• Start with the smallest:
1 | 1 -3 17 16 20
1 -2 15 31
1 -2 15 31 51
2 | 1 -3 17 16 20
2 -2 30 92
1 -1 15 46 112
48
Continued
4 | 1 -3 17 16 20
4 4 84 400
1 1 21 100 420 Since all are positive, 4 is an upper bound
Remainders are getting larger. Let’s try negatives
-1 | 1 -3 17 16 20
-1 4 -21 5
1 -4 21 -5 15 Note that they alternate sign,
So -1 is an lower bound
Therefore, the equation has no positive or negative rational
roots. Zero is not a root, so there are no rational roots
49
Continued
• ± 1, 2, 4, 5, 10, 20, so are the possible roots
• Roots must be ≤4 and ≥-1, leaving only -1, 1, 2 and 4 as
possible roots
• In getting to the bounds, we showed that none of these are
root, which means that there are no rational roots to this
equation
50
Example
• Find the rational roots 2x3 – 5x2 – 3x + 9 = 0
51
Solution
• Find the rational roots 2x3 – 5x2 – 3x + 9 = 0
• Possible roots are ±1, 3, 9, ½, 3/2, 9/2
1| 2 -5 -3 9
-3| 2 -5 -3 8
2 -3 -6
-6 33 -30
2 -3 -6 3
2 -11 30 -22 alternate, -3 is lower bound
-1| 2 -5 -3 9
-2 7 -4
2 -7 4 5
3| 2 -5 -3 9
6 3 0
2 1 0 9 all positive, so 3 is an upper bound
52
continued
Know roots are between -3 and 3
1/2| 2 -5 -3 9
1 -2 -5/2
2 -4 -5
3/2| 2 -5 -3 9 3/2 is a root!
3 -3 -9
2 -2 -6 0
(2x3
–
5x2
– 3x + 9)/(x-3/2) =
2x2
-2x -6,
2± 22 +4∗12
x=
4
=
1± 13
2
53
Example
• Find the roots of x3 + 2x2 – 3x +20 = 0
54
Example
• Find the roots of x3 + 2x2 – 3x +20 = 0
• Possible roots are: ± 1, 2, 4, 5, 10, 20
1| 1 2 -3 20
2 4 1
1 4 1 21 all positive, so 1 is an upper bound
-1| 1 2 -3 20
-2| 1 2 -3 20
-1 -1 4
-2 0 6
1 1 -4 38
1 0 -3 24
55
Solution cont
• Find the roots of x3 + 2x2 – 3x +20 = 0
• Possible roots are: ± 1, 2, 4, 5, 10, 20
-4| 1 2 -3
20
-8 20 -68
1 -6 17 -48 alternating, so is lower bound
Roots are bounded by 1 and -4, but none work. No rational roots
56
Location Theorem
• If f(a)< 0 and f(b)>0, then for some x, a < x < b, f(x)=0
• In other words, the curve has to cross the axis (=0) to change
its sign
• How do we use this? Try numbers, see when the value
changes sign
57
Example
• f(x) =x5 – 3x2 + 4x + 2= 0
•
This has a root between -1 and 0
(f(0) = 2, f(-1) = -1- 3 - 4 + 2 = - 6; sign changes)
• Approximate the root
58
Solution
• f(x) =x5 – 3x2 + 4x + 2= 0
• Try -½: 2 – 2 -3/4 - 1/128 <0, needs to be larger
• Try -1/4: 2 – 1 – 3/16 -1/1024>0, needs to be smaller
so root is between -1/4 and -1/2
59
Examples
• Show these equations have no rational roots
x3 + 8x – 1 = 0
x4 + 4x3 + 4x2 – 16 = 0
60
Solution
•
x3 + 8x – 1 = 0
• Possible roots are 1, -1, neither one works
61
Solution
• x4 + 4x3 + 4x2 – 16 = 0
• Possible roots are 1, 2, 4, 16
• Let’s simplify: x4 + 4(x3 + x2 – 4)=0; only negative is -4, so
x3+x2 have to be <4 for it to work. This can’t happen for
anything but 1, -1
(x3 + x2 – 4) is -4 for both 1 and -1; the second term is -16,
and 1-16  0
62
Solve
• 2x3 – 5x2 – 3x + 9 = 0
63
Solution
• 2x3 – 5x2 – 3x + 9 = 0
• Possible roots are  1, 3, 9 divided by  1, 2
1 | 2 -5 -3 9
3 | 2 -5 -3 9
2 -3 -18
6 -3 -18
2 -3 -6 -9
2 -1 -6 -9
9/2 | 2 -5 -3 9
9 18
2
4
15 >0
9/2 is upper bound, 3 is not – don’t have to try 3/2 or 9
-1| 2 -5 -3 9
-2 7 -4
2 -7 4 5
-3| 2 -5 -3 9
-3/2| 2 -5 -3 9
-6 33 -99
2 -11 30 -90
-3 12 -27/2
2 -8 9 <0
-3/2 is a lower bound; nothing between them works!
64
Chapter 13.5
Conjugate Rule of Signs
65
Conjugates
• When we are dealing with complex numbers as roots to
quadratics, the come in pairs
a+ ib and a – ib
• It turns out that this is true for all polynomials with real
coefficients;
if a+ ib is a root to an equation then so is a – ib
• a+ ib and a – ib are called conjugates
66
A few rules about conjugates
• If z1 and z2 are complex numbers, and the conjugate of z is z:
z1 z2 = z1z2
(z1)n= z1n
r = r if r is real
z1 + z2 = z1 + z2
67
Example
• If 1 + 3i is a root of f(x) = 2x4 – 3x3 + 12x2 + 22x – 60 = 0, find
the other roots
68
Solution
• If 1 + 3i is a root of f(x) = 2x4 – 3x3 + 12x2 + 22x – 60 = 0, find
the other roots
• We know 1 - 3i is also a root
(x -1-3i) (x -1 + 3i) =(x2 –2x + 10) is a factor
2x2 + x + 6
• x2 –2x + 10 | 2x4 – 3x3 + 12x2 + 22x – 60
2x4 - 4x3 +20x2
x3 -8x2 + 22x
x3 - 2x2 + 10x
6x2 + 12x – 60
6x2 + 12x - 60
69
Solution, contd
• Roots of 2x2 + x + 6 are x = 3/2, x = -2
70
Example
• Show that x3 – 2x2 + x – 1=0 has at least one irrational root
71
Solution
• Show that x3 – 2x2 + x – 1=0 has at least one irrational root
• Possible rational roots are 1, -1
Neither one works, so no rational roots
• Complex roots are in pairs, but there are 3 solutions to this
equations
• Therefore, one of the roots must be irrational
72
Irrational Roots
• If all the coefficients of a polynomial are rational, and
a + b 𝑐 is a root, the a - b 𝑐 is a root;
• Irrational roots also come in pairs!
73
Example
• Find a polynomial with rational coefficients with the root
4+5 3
74
Solution
• Find a polynomial with rational coefficients with the root
4+5 3
• If 4 + 5 3 is a root, the so is 4 - 5 3
• (x - 4 + 5 3)(x - 4 - 5 3) = x2 – 89x - 59
75
Descartes Rule of Signs
• If f(x) is a polynomial with real coefficients, and f(x)=0 then:
– the number of positive roots is equal to the variations in
sign of x, or less that by an even integer
– The number of negative roots is equal to the variation in
sign of f(-x) or less than that by an even integer
76
Example
• Investigate the roots of x3 + 8x + 5 = 0
There are no variation in signs, thus there are 0 positive roots
• f(-x) = -x3 - 8x + 5 = 0
There is one sign change, so there is 1 negative root
• Therefore, there is one negative root and no positive roots
• Possible rational roots, that are negative, or -1 and -5
(-1) +8(-1) + 5  0, (-125) + 8(-5) + 5  0, so there are no
rational roots
• Therefore, there is one, negative real root and two (non-real)
complex roots
77
Example
• Investigate the roots of x4 + 3x2 – 7x – 5 = 0
78
Solution
• Investigate the roots of x4 + 3x2 – 7x – 5 = 0
– There is one sign variation, so one positive root
• f(-x) = x4 + 3x2 + 7x – 5 = 0
– There is one sign variation, so one negative root
• Since there are 4 roots, there are two complex roots
• Rational roots can be 1, -1, 5, -5
1 + 3 – 7 + 5  0, 1 + 3 + 7 – 5  0
625 + 75 – 35 – 5  0
625 + 74 +35 - 5  0; no rational roots
79
Example
• Investigate the roots of x3 – x2 + 3x + 2 = 0
– There are two variations of sign, so there are two positive
roots or zero positive roots
• f(-x) = -x3 – x2 - 3x + 2 = 0
– There is one variation in sign, so we have one
negative root
• This allows two possibilities:
– There are two positive roots and one negative root
– There is one negative root and two complex roots
80
Example
• If x=4 + i, find the remaining roots of
x4 - 10x3 + 30x2 – 10x – 51=0
81
Solution
• If x=4 + i, find the remaining roots of
x4 - 10x3 + 30x2 – 10x – 51=0
• (x – 4 - i )(x – 4 +i) = x2 – 8x + 17
x2 - 2x -3
x2 – 8x + 17 | x4 - 10x3 + 30x2 – 10x – 51
x4 -8x3 + 17x2
-2x3 + 13 x2 - 10x
-2x3 + 16x2 - 34x
-3x2 - 24x - 51
-3x2 - 24x -51
x2 - 2x -3=(x-3)(x+1), x=3, -1, 4-i
82
Example
• Use Descartes’ rule of signs to find information about the
roots of: x4 + x2 + 1 = 0
83
Solution
• Use Descartes’ rule of signs to find information about the
roots of: x4 + x2 + 1 = 0
• No changes in sign, so no positive roots
• Subst. –x, no sign changes, so no negative roots
• Only roots are complex, four of them
84
Example
• Use Descartes’ rule of signs to find information about the
roots of: x3 - 4x2 + x - 1 = 0
85
Solution
• Use Descartes’ rule of signs to find information about the
roots of: x3 - 4x2 + x - 1 = 0
3 changes of sign, so 3 or 1 positive roots
• Subst. –x: -x3 - 4x2 - x - 1 = 0
No sign changes, so no negative roots
Either 3 positive roots, or one positive root and a complex pair
86
Chapter 13.6
Introduction to Partial Fractions
87
Partial Fractions
• For example, rewrite:
1
𝑥 2 −1
• Find A and B such that
=
1
2(𝑥−1)
2𝑥 −3
(𝑥−1)(𝑥+1)
-
=
1
2(𝑥+1)
𝐴
𝑥−1
+
𝐵
𝑥+1
88
Solution
• Find A and B such that
2𝑥 −3
(𝑥−1)(𝑥+1)
=
𝐴
𝑥−1
+
𝐵
𝑥+1
• 2x-3 = A(x+1) + B(x-1)
• A+B=2, A-B=-3
• A + -1/2, B = 5/2
89
Example
• Find A and B:
𝑥
(𝑥+4)2
=
𝐴
𝑥+4
+
𝑩
(𝑥+4)2
90
Solution
• Find A and B:
𝑥
(𝑥+4)2
=
𝐴
𝑥+4
+
𝑩
(𝑥+4)2
• x = A(x+4) + B
• A = 1, 4A = -B = -4
91
Alternate Method
• Find A and B:
𝑥
(𝑥+4)2
=
𝐴
𝑥+4
+
𝑩
(𝑥+4)2
• x = A(x+4) + B
• Let x = -4, B = -4
• Then use any value for x, x = 0, and get 4A + B = 0, A = 1
92
Example: Use Both Methods
• Find A, B, and C:
7 𝑥 2 −9𝑥+29
(𝑥−2)(𝑥 2 +9)
=
𝐴
𝑥−2
+
𝐵𝑥+𝐶
𝑥 2 +9
93
Solution, direct method
• Find A, B, and C:
7 𝑥 2 −9𝑥+29
(𝑥−2)(𝑥 2 +9)
=
𝐴
𝑥−2
+
𝐵𝑥+𝐶
𝑥 2 +9
• 7x2 – 9x + 29 = A(x2+9) + (Bx+C)(x-2)
7 = A+B
-9 = -2B +C
29= 9A – 2C
A = 3, B = 4, C = -1
94
Alternate Method
• Find A, B, and C:
7 𝑥 2 −9𝑥+29
(𝑥−2)(𝑥 2 +9)
=
𝐴
𝑥−2
+
𝐵𝑥+𝐶
𝑥 2 +9
• 7x2 – 9x + 29 = A(x2+9) + (Bx+C)(x-2)
Let x = 2, 28 – 18 + 29 = A(13), 39=13A, A = 3
Let x = 3i, -63 – 27i + 29 = (B(3i) + C)(3i – 2)
34 – 27i = 3i(2B + C) – 2C – 9B
34 = -2C – 9B, 27 = 6B + 3C or 9 = 2B + C
B = 4, C = -1
95