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8-6
Theorems About Roots of Polynomial
Equations
TEKS FOCUS
VOCABULARY
ĚConjugates – number pairs of the form
TEKS (7)(E) Determine linear and quadratic factors of a polynomial
expression of degree three and of degree four, including factoring the
sum and difference of two cubes and factoring by grouping.
a + 1b and a - 1b
ĚRepresentation – a way to display or
TEKS (1)(E) Create and use representations to organize, record, and
communicate mathematical ideas.
describe information. You can use a
representation to present mathematical
ideas and data.
Additional TEKS (1)(F), (7)(A), (7)(D)
ESSENTIAL UNDERSTANDING
The factors of the numbers an and a0 in P (x) = anx n + an-1x n-1 + c + a1x + a0
can help you factor P (x) and solve the equation P (x) = 0.
Theorem
Rational Root Theorem
Let P (x) = anx n + an-1x n-1 + c + a1x + a0 be a polynomial with integer
coefficients. There are a limited number of possible roots of P (x) = 0:
r Integer roots must be factors of a0 .
p
r Rational roots must have reduced form q where p is an integer factor of a0 and q is
an integer factor of an.
Factors of the leading
coefficient:
1, 3, 7, and 21.
21x2 + 29x + 10 = 0
x2
29
10
+ 21x + 21 = 0
( x + )( x + ) = 0
2
3
Theorem
5
7
Factors of the constant term:
1, 2, 5, and 10.
The roots are 23 and 57.
Conjugate Root Theorem
If P (x) is a polynomial with rational coefficients, then irrational roots of P (x) = 0
that have the form a + 1b occur in conjugate pairs. That is, if a + 1b is an
irrational root with a and b rational, then a - 1b is also a root.
If P (x) is a polynomial with real coefficients, then the complex roots of P (x) = 0
occur in conjugate pairs. That is, if a + bi is a complex root with a and b real,
then a - bi is also a root.
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369
Theorem
Descartes’ Rule of Signs
Let P (x) be a polynomial with real coefficients written in standard form.
r The number of positive real roots of P (x) = 0 is either equal to the number of sign
changes between consecutive coefficients of P (x) or is less than that by an even
number.
r The number of negative real roots of P (x) = 0 is either equal to the number of sign
changes between consecutive coefficients of P ( -x) or is less than that by an even
number.
In both cases, count multiple roots according to their multiplicity.
Problem 1
P
Finding a Rational Root
What are the rational roots of 2x 3 − x 2 + 2x + 5 = 0?
W
What information
can you get from the
equation?
7KHHTXDWLRQJLYHV\RX
WKHOHDGLQJFRHIILFLHQW
DQGWKHFRQVWDQWWHUP
factor of constant term
T only possible rational roots have the form factor
The
of leading coefficient .
T constant factors are {1, {5. The leading coefficient factors are {1, {2.
The
T only possible rational roots are {1, {5, { 12, { 52 .
The
The table shows the values of the function y = P (x) for the possible roots.
x
1
1
5
5
1
2
12
P(x)
8
0
240
280
6
7
2
5
2
52
35
75
2
The only rational root of 2x 3 - x 2 + 2x + 5 = 0 is -1.
Problem
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2
TEKS Process Standard (1)(E)
Using the Rational Root Theorem
What are the rational roots of 15x 3 − 32x 2 + 3x + 2 = 0?
Coefficients and the
constant term of
the polynomial
The roots of
the polynomial
equation
Ě)LQGRQHURRW
Ě)DFWRUXQWLO\RXJHWDTXDGUDWLF
Ě8VHTXDGUDWLFHTXDWLRQVROXWLRQ
PHWKRGVWRILQGWKHRWKHUURRWV
Step 1 The constant term factors are {1 and {2. The leading coefficient factors
are {1, {3, {5, and {15.
continued on next page ▶
370
Lesson 8-6
Theorems About Roots of Polynomial Equations
Problem 2
continued
1
2
Step 2 The possible rational roots are: {1, {2, { 13, { 23, { 15, { 25, { 15
, and { 15
.
Step 3 Test each possible rational root in 15x 3 - 32x 2 + 3x + 2 until you
find a root.
Test 1: 15(1) 3 - 32(1) 2 + 3(1) + 2 = -12 ≠ 0
Test 2: 15(2) 3 - 32(2) 2 + 3(2) + 2 = 0
6RLVDURRW
Step 4 Factor the polynomial by using synthetic division: P (x) = (x - 2) 1 15x 2 - 2x - 1 2 .
2
15
15
-32
30
-2
3
-4
-1
2
-2
0
Step 5 Since 15x 2 - 2x - 1 = (5x + 1)(3x - 1), the other roots are - 15 and 13 .
The rational roots of 15x 3 - 32x 2 + 3x + 2 = 0 are 2, - 15, and 13.
Problem
bl
3
Using the Conjugate Root Theorem to Identify Roots
Do you have real
coefficients?
All rational numbers are
UHDOQXPEHUV7KHUHIRUH
the rational coefficients
DUHUHDOFRHIILFLHQWV
A quartic polynomial P (x) has rational coefficients. If 12 and 1 + i are roots of
P (x) = 0, what are the two other roots?
S
Since P (x) has rational coefficients and 0 + 12 is a root of P (x) = 0, it follows from
tthe Conjugate Root Theorem that 0 - 12 is also a root.
S
Since
P (x) has real coefficients and 1 + i is a root of P (x) = 0, it follows that 1 - i is
aalso a root.
The two other roots are - 12 and 1 - i.
Problem
bl
4
Using Conjugates to Construct a Polynomial
Multiple Choice What is a third-degree polynomial function y = P (x) with
rational coefficients so that P (x) = 0 has roots −4 and 2i?
Does the Conjugate
Root Theorem apply
to −4?
No; the theorem does
not apply because -4
is neither irrational nor
LPDJLQDU\
P (x) = x 3 - 2x 2 - 16x + 32
P (x) = x 3 + 4x 2 + 4x + 16
P (x) = x 3 - 4x 2 + 4x - 16
P (x) = x 3 + 4x 2 - 4x - 16
S
Since 2i is a root, then -2i is also a root.
P (x) = (x + 2i)(x - 2i)(x + 4)
=
(x 2
+ 4)(x + 4)
= x 3 + 4x 2 + 4x + 16
x3
:ULWHWKHSRO\QRPLDOIXQFWLRQ
0XOWLSO\WKHFRPSOH[FRQMXJDWHV
:ULWHWKHSRO\QRPLDOIXQFWLRQLQVWDQGDUGIRUP
The equation
+
+ 4x + 16 = 0 has rational coefficients and has roots -4
and 2i. The correct answer is C.
4x 2
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Problem 5
P
TEKS Process Standard (1)(F)
Using Descartes’ Rule of Signs
What does Descartes’ Rule of Signs tell you about the real roots of
x 3 − x 2 + 1 = 0?
There
are two sign changes, + to - and - to + .
T
Therefore,
there are either 0 or 2 positive real roots.
T
P ( -x) = ( -x)3 - ( -x)2 + 1 = -x 3 - x 2 + 1 = 0 has only one sign change, - to + .
There is one negative real root.
T
Recall that graphs of cubic functions have zero or two turning points. Because the
R
ggraph already shows two turning points, it will not change direction again. So there
are no positive real roots.
NLINE
HO
ME
RK
O
Why can’t there be
zero negative real
roots?
The number of negative
roots is equal to 1 or is
less than 1 by an even
number. Zero is less than
1 by an odd number.
WO
PRACTICE and APPLICATION EXERCISES
Scan page for a Virtual Nerd™ tutorial video.
Use the Rational Root Theorem to list all possible rational roots for each equation.
Then find any actual rational roots.
For additional support when
completing your homework,
go to PearsonTEXAS.com.
1. x 3 - 4x + 1 = 0
2. x 3 + 2x - 9 = 0
3. 2x 3 - 5x + 4 = 0
4. 3x 3 + 9x - 6 = 0
5. 4x 3 + 2x - 12 = 0
6. 6x 3 + 2x - 18 = 0
7. 7x 3 - x 2 + 4x + 10 = 0
8. 8x 3 + 2x 2 - 5x + 1 = 0
9. 10x 3 - 7x 2 + x - 10 = 0
10. Apply Mathematics (1)(A) You are building a square pyramid out of clay and
want the height to be 0.5 cm shorter than twice the length of each side of the
base. If you have 18 cm3 of clay, what is the greatest height you could use for
your pyramid?
372
Lesson 8-6
Theorems About Roots of Polynomial Equations
A polynomial function P (x) with rational coefficients has the given roots. Find two
additional roots of P (x) = 0.
11. -2i and 110
12. 14 - 12 and -6i
13. i and 7 + 8i
14. - 13 and 5 - 111
Write a polynomial function with rational coefficients so that P (x) = 0 has the
given roots.
15. 7 and 12
16. -9 and -15
17. -10i
18. 3i + 9
19. 4, 16, and 1 + 19i
20. 13i and 5 + 10i
21. 11 - 2i and 8 + 13i
22. 17 - 4i and 12 + 5i
What does Descartes’ Rule of Signs say about the number of positive real roots
and negative real roots for each polynomial function?
23. P (x) = x 2 + 5x + 6
24. P (x) = 9x 3 - 4x 2 + 10 25. P (x) = 8x 3 + 2x 2 - 14x + 5
26. Evaluate Reasonableness (1)(B) Your friend is using
Descartes’ Rule of Signs to find the number of negative
real roots of x 3 + x 2 + x + 1 = 0. Describe and correct
the error.
P(–x) = (–x)3 + (–x)2 + (–x) + 1
= –x3 – x2 – x + 1
Because there is only one sign
change in P(–x), there must be one
negative real root.
27. Explain Mathematical Ideas (1)(G) A quartic equation
with integer coefficients has two real roots and one
imaginary root. Explain why the fourth root must be
imaginary.
28. Apply Mathematics (1)(A) A climbing gym is designing a new bouldering section
in the shape of a trapezoid. They want the shorter base to be twice the height and the
longer base to be 4 feet longer than the shorter base. If they have enough shredded
rubber to create a 60 ft2 bouldering area, what dimensions should they use in the gym?
2h
h
2h + 4
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Find all rational roots for P (x) = 0.
29. P (x) = 2x 3 - 5x 2 + x - 1
30. P (x) = 6x 4 - 13x 3 + 13x 2 - 39x - 15
31. P (x) = 7x 3 - x 2 - 5x + 14
32. P (x) = 3x 4 - 7x 3 + 10x 2 - x + 12
33. P (x) = 6x 4 - 7x 2 - 3
34. P (x) = 2x 3 - 3x 2 - 8x + 12
35. Connect Mathematical Ideas (1)(F) Write a fourth-degree polynomial equation
with integer coefficients that has two irrational roots and two imaginary roots.
36. a. Find a polynomial equation in which 1 + 12 is the only root.
b. Find a polynomial equation with root 1 + 12 of multiplicity 2.
c. Find c such that 1 + 12 is a root of x 2 - 2x + c = 0.
37. a. Using real and imaginary as types of roots, list all possible combinations of
root type for a fourth-degree polynomial equation.
b. Repeat the process for a fifth-degree polynomial equation.
c. Make a conjecture about the number of real roots of an odd-degree
polynomial equation.
38. Justify Mathematical Arguments (1)(G) A student states that 2 + 13 is a root
of x 2 - 2x - (3 + 213) = 0. The student claims that 2 - 13 is another root of
the equation by the Conjugate Root Theorem. Explain how you would respond
to the student.
TEXAS Test Practice
T
39. What is a positive root of -5x3 - 2x2 + 9x + 30 = 0?
40. What is the remainder when you divide x3 + 2x2 - x - 6 by x - 1?
41. A polynomial with rational coefficients has roots -3i and 8 + 27. What is the
minimum degree of the polynomial?
42. What is the value of y in the solution of the system of equations below?
= 9
{ 10x8x ++ 24y
60y = 14
43. What is the value of the greater solution of the equation 6x2 - 17x + 5 = 0?
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Lesson 8-6 Theorems About Roots of Polynomial Equations