The document discusses three theorems for finding real roots of real polynomials: 1) Descartes' Rule of Signs determines the possible number of positive or negative real roots based on the variation in signs of the polynomial's coefficients. 2) The Theorem on the Bounds gives the interval where real roots must reside. 3) The Theorem on Rational Roots identifies possible rational roots for polynomials with integer coefficients.

1. Looking for Real Roots of Real Polynomials

2. In this section we give three theorems concerning
real roots of real polynomials.
Looking for Real Roots of Real Polynomials

3. In this section we give three theorems concerning
real roots of real polynomials. They give three
different approaches concerning looking for roots.
Looking for Real Roots of Real Polynomials

4. In this section we give three theorems concerning
real roots of real polynomials. They give three
different approaches concerning looking for roots.
Looking for Real Roots of Real Polynomials
Descartes' Rule of Signs gives the possible
number of real roots by eyeballing the polynomial.

5. In this section we give three theorems concerning
real roots of real polynomials. They give three
different approaches concerning looking for roots.
Looking for Real Roots of Real Polynomials
Descartes' Rule of Signs gives the possible
number of real roots by eyeballing the polynomial.
Theorem on the Bounds gives the interval of real
numbers where the roots must reside.

6. In this section we give three theorems concerning
real roots of real polynomials. They give three
different approaches concerning looking for roots.
Looking for Real Roots of Real Polynomials
Descartes' Rule of Signs gives the possible
number of real roots by eyeballing the polynomial.
Theorem on the Bounds gives the interval of real
numbers where the roots must reside.
Theorem on Rational Roots gives the possible
rational roots for polynomials with integer coefficients.

7. In this section we give three theorems concerning
real roots of real polynomials. They give three
different approaches concerning looking for roots.
Looking for Real Roots of Real Polynomials
Descartes' Rule of Signs gives the possible
number of real roots by eyeballing the polynomial.
Theorem on the Bounds gives the interval of real
numbers where the roots must reside.
Descartes' Rule and Theorem on the Bounds
are existence-theorems in mathematics,
i.e. they establish the existence of something but
don’t say what they might be.
Theorem on Rational Roots gives the possible
rational roots for polynomials with integer coefficients.

8. In this section we give three theorems concerning
real roots of real polynomials. They give three
different approaches concerning looking for roots.
Looking for Real Roots of Real Polynomials
Descartes' Rule of Signs gives the possible
number of real roots by eyeballing the polynomial.
Theorem on the Bounds gives the interval of real
numbers where the roots must reside.
Descartes' Rule and Theorem on the Bounds
are existence-theorems in mathematics,
i.e. they establish the existence of something but
don’t say what they might be. Theorem on Rational
Roots tells us precisely what to check.
Theorem on Rational Roots gives the possible
rational roots for polynomials with integer coefficients.

9. Looking for Real Roots of Real Polynomials
Theorem: If the degree of a real polynomial P(x)
is odd then P(x) must have an odd number of real
roots.

10. Looking for Real Roots of Real Polynomials
Theorem: If the degree of a real polynomial P(x)
is odd then P(x) must have an odd number of real
roots. In particular, it has at least one real root.

11. Looking for Real Roots of Real Polynomials
Theorem: If the degree of a real polynomial P(x)
is odd then P(x) must have an odd number of real
roots. In particular, it has at least one real root.
Proof:
From the Fundamental Theorem Algebra the
complex roots must be in conjugate pairs.

12. Looking for Real Roots of Real Polynomials
Theorem: If the degree of a real polynomial P(x)
is odd then P(x) must have an odd number of real
roots. In particular, it has at least one real root.
Proof:
From the Fundamental Theorem Algebra the
complex roots must be in conjugate pairs.
Hence there must be even number of complex roots.

13. Looking for Real Roots of Real Polynomials
Theorem: If the degree of a real polynomial P(x)
is odd then P(x) must have an odd number of real
roots. In particular, it has at least one real root.
Proof:
From the Fundamental Theorem Algebra the
complex roots must be in conjugate pairs.
Hence there must be even number of complex roots.
So the remaining roots must be real and there must
be an odd number of them. QED

14. Looking for Real Roots of Real Polynomials
Theorem: If the degree of a real polynomial P(x)
is odd then P(x) must have an odd number of real
roots. In particular, it has at least one real root.
Proof:
Given a polynomial P(x), arrange the signs of it's
coefficients starting from the highest degree term in
descending order,
From the Fundamental Theorem Algebra the
complex roots must be in conjugate pairs.
Hence there must be even number of complex roots.
So the remaining roots must be real and there must
be an odd number of them. QED

15. Looking for Real Roots of Real Polynomials
Theorem: If the degree of a real polynomial P(x)
is odd then P(x) must have an odd number of real
roots. In particular, it has at least one real root.
Proof:
Given a polynomial P(x), arrange the signs of it's
coefficients starting from the highest degree term in
descending order, the total number of sign-changes
between consecutive terms is called
the variation of the signs of P(x).
From the Fundamental Theorem Algebra the
complex roots must be in conjugate pairs.
Hence there must be even number of complex roots.
So the remaining roots must be real and there must
be an odd number of them. QED

16. Looking for Real Roots of Real Polynomials
For example, if P(x) = –3x5 + x3 + 2x2 + x – 1,
the signs of its coefficients in order are
– + + + –

17. Looking for Real Roots of Real Polynomials
For example, if P(x) = –3x5 + x3 + 2x2 + x – 1,
the signs of its coefficients in order are
– + + + –
sign switched sign switched

18. Looking for Real Roots of Real Polynomials
For example, if P(x) = –3x5 + x3 + 2x2 + x – 1,
the signs of its coefficients in order are
– + + + –
There are two sign-switches so the variation of Signs
of P(x) is 2.
sign switched sign switched

19. Looking for Real Roots of Real Polynomials
For example, if P(x) = –3x5 + x3 + 2x2 + x – 1,
the signs of its coefficients in order are
– + + + –
There are two sign-switches so the variation of Signs
of P(x) is 2.
sign switched sign switched
Descartes' Rule of Signs:

20. Looking for Real Roots of Real Polynomials
For example, if P(x) = –3x5 + x3 + 2x2 + x – 1,
the signs of its coefficients in order are
– + + + –
There are two sign-switches so the variation of Signs
of P(x) is 2.
sign switched sign switched
Descartes' Rule of Signs: P(x) is a real
polynomial,

21. Looking for Real Roots of Real Polynomials
For example, if P(x) = –3x5 + x3 + 2x2 + x – 1,
the signs of its coefficients in order are
– + + + –
There are two sign-switches so the variation of Signs
of P(x) is 2.
sign switched sign switched
Descartes' Rule of Signs: P(x) is a real polynomial,
a. the number of positive roots of P(x) is equal to the
variation of Signs of P(x) or less than the variation of
Signs of P(x) by an even number.

22. Looking for Real Roots of Real Polynomials
For example, if P(x) = –3x5 + x3 + 2x2 + x – 1,
the signs of its coefficients in order are
– + + + –
There are two sign-switches so the variation of Signs
of P(x) is 2.
sign switched sign switched
Descartes' Rule of Signs: P(x) is a real polynomial,
a. the number of positive roots of P(x) is equal to the
variation of Signs of P(x) or less than the variation of
Signs of P(x) by an even number.
b. the number of negative roots of P(x) is equal to the
variation of Signs of P(–x) or less than the variation
of Signs of P(–x) by an even number.

23. Looking for Real Roots of Real Polynomials
Example A. For P(x) = –3x5 + x3 + 2x2 + x – 1 from
the last slide, the variation of signs is 2.

24. Looking for Real Roots of Real Polynomials
Example A. For P(x) = –3x5 + x3 + 2x2 + x – 1 from
the last slide, the variation of signs is 2. Hence P(x)
has either 2 positive roots or no positive roots.

25. Looking for Real Roots of Real Polynomials
Example A. For P(x) = –3x5 + x3 + 2x2 + x – 1 from
the last slide, the variation of signs is 2. Hence P(x)
has either 2 positive roots or no positive roots.
P(-x) = -3(-x)5 + (-x)3 + 2(-x)2 + (-x) – 1

26. Looking for Real Roots of Real Polynomials
Example A. For P(x) = –3x5 + x3 + 2x2 + x – 1 from
the last slide, the variation of signs is 2. Hence P(x)
has either 2 positive roots or no positive roots.
P(-x) = -3(-x)5 + (-x)3 + 2(-x)2 + (-x) – 1
= 3x5 – x3 + 2x2 – x – 1

27. Looking for Real Roots of Real Polynomials
Example A. For P(x) = –3x5 + x3 + 2x2 + x – 1 from
the last slide, the variation of signs is 2. Hence P(x)
has either 2 positive roots or no positive roots.
P(-x) = -3(-x)5 + (-x)3 + 2(-x)2 + (-x) – 1
= 3x5 – x3 + 2x2 – x – 1
The signs of its coefficients in order are
+ – + – –

28. Looking for Real Roots of Real Polynomials
Example A. For P(x) = –3x5 + x3 + 2x2 + x – 1 from
the last slide, the variation of signs is 2. Hence P(x)
has either 2 positive roots or no positive roots.
P(-x) = -3(-x)5 + (-x)3 + 2(-x)2 + (-x) – 1
= 3x5 – x3 + 2x2 – x – 1
The signs of its coefficients in order are
+ – + – –
sign switched

29. Looking for Real Roots of Real Polynomials
Example A. For P(x) = –3x5 + x3 + 2x2 + x – 1 from
the last slide, the variation of signs is 2. Hence P(x)
has either 2 positive roots or no positive roots.
P(-x) = -3(-x)5 + (-x)3 + 2(-x)2 + (-x) – 1
= 3x5 – x3 + 2x2 – x – 1
The signs of its coefficients in order are
+ – + – –
sign switched
the variation of signs of P(-x) is 3.

30. Looking for Real Roots of Real Polynomials
Example A. For P(x) = –3x5 + x3 + 2x2 + x – 1 from
the last slide, the variation of signs is 2. Hence P(x)
has either 2 positive roots or no positive roots.
P(-x) = -3(-x)5 + (-x)3 + 2(-x)2 + (-x) – 1
= 3x5 – x3 + 2x2 – x – 1
The signs of its coefficients in order are
+ – + – –
sign switched
the variation of signs of P(-x) is 3. Hence P(x) has
either 3 negative roots or 1 negative root.

31. Looking for Real Roots of Real Polynomials
Example A. For P(x) = –3x5 + x3 + 2x2 + x – 1 from
the last slide, the variation of signs is 2. Hence P(x)
has either 2 positive roots or no positive roots.
P(-x) = -3(-x)5 + (-x)3 + 2(-x)2 + (-x) – 1
= 3x5 – x3 + 2x2 – x – 1
The signs of its coefficients in order are
+ – + – –
sign switched
the variation of signs of P(-x) is 3. Hence P(x) has
either 3 negative roots or 1 negative root.
So P(x) may have 1 neg. root + 4 complex roots,
or 1 neg. root + 2 positive roots + 2 complex roots,
or 3 negative roots + 2 complex roots,
or 3 negative roots + 2 positive roots.

32. Looking for Real Roots of Real Polynomials
The Theorem of Bounds gives the interval of real
numbers where the roots must reside.

33. Theorem of Bounds:
Looking for Real Roots of Real Polynomials
The Theorem of Bounds gives the interval of real
numbers where the roots must reside.

34. Theorem of Bounds:
Let P(x) = anxn + an-1xn-1+… + a0 be a real polynomial,
Looking for Real Roots of Real Polynomials
The Theorem of Bounds gives the interval of real
numbers where the roots must reside.

35. Theorem of Bounds:
Let P(x) = anxn + an-1xn-1+… + a0 be a real polynomial,
the real roots of P(x) must be in the interval (–M, M)
where M =
Looking for Real Roots of Real Polynomials
The Theorem of Bounds gives the interval of real
numbers where the roots must reside.
Max {|an|, |an-1|, |an-2|, ..|a0|}
|an|
+ 1

36. Theorem of Bounds:
Let P(x) = anxn + an-1xn-1+… + a0 be a real polynomial,
the real roots of P(x) must be in the interval (–M, M)
where M =
Looking for Real Roots of Real Polynomials
The Theorem of Bounds gives the interval of real
numbers where the roots must reside.
Max {|an|, |an-1|, |an-2|, ..|a0|}
|an|
+ 1
Example B. For P(x) = -2x5 + 6x3 + 2x2 + x – 1,

37. Theorem of Bounds:
Let P(x) = anxn + an-1xn-1+… + a0 be a real polynomial,
the real roots of P(x) must be in the interval (–M, M)
where M =
Looking for Real Roots of Real Polynomials
The Theorem of Bounds gives the interval of real
numbers where the roots must reside.
Max {|an|, |an-1|, |an-2|, ..|a0|}
|an|
+ 1
Example B. For P(x) = -2x5 + 6x3 + 2x2 + x – 1,
then Max{|-2|, |6|, |2|, |1|, |-1|} = 6

38. Theorem of Bounds:
Let P(x) = anxn + an-1xn-1+… + a0 be a real polynomial,
the real roots of P(x) must be in the interval (–M, M)
where M =
Looking for Real Roots of Real Polynomials
The Theorem of Bounds gives the interval of real
numbers where the roots must reside.
Max {|an|, |an-1|, |an-2|, ..|a0|}
|an|
+ 1
Example B. For P(x) = -2x5 + 6x3 + 2x2 + x – 1,
then Max{|-2|, |6|, |2|, |1|, |-1|} = 6 and |an| = 2,

39. Theorem of Bounds:
Let P(x) = anxn + an-1xn-1+… + a0 be a real polynomial,
the real roots of P(x) must be in the interval (–M, M)
where M =
Looking for Real Roots of Real Polynomials
The Theorem of Bounds gives the interval of real
numbers where the roots must reside.
Max {|an|, |an-1|, |an-2|, ..|a0|}
|an|
+ 1
Example B. For P(x) = -2x5 + 6x3 + 2x2 + x – 1,
then Max{|-2|, |6|, |2|, |1|, |-1|} = 6 and |an| = 2,
hence M = 6/2 + 1 = 4 and all the real roots of P(x)
reside in the interval (–4, 4).

40. Theorem of Bounds:
Let P(x) = anxn + an-1xn-1+… + a0 be a real polynomial,
the real roots of P(x) must be in the interval (–M, M)
where M =
Looking for Real Roots of Real Polynomials
The Theorem of Bounds gives the interval of real
numbers where the roots must reside.
Max {|an|, |an-1|, |an-2|, ..|a0|}
|an|
+ 1
Example B. For P(x) = -2x5 + 6x3 + 2x2 + x – 1,
then Max{|-2|, |6|, |2|, |1|, |-1|} = 6 and |an| = 2,
hence M = 6/2 + 1 = 4 and all the real roots of P(x)
reside in the interval (–4, 4).
Note: If an = 1, then M = largest Coefficient + 1.

41. Theorem of Bounds:
Let P(x) = anxn + an-1xn-1+… + a0 be a real polynomial,
the real roots of P(x) must be in the interval (–M, M)
where M =
Looking for Real Roots of Real Polynomials
The Theorem of Bounds gives the interval of real
numbers where the roots must reside.
Max {|an|, |an-1|, |an-2|, ..|a0|}
|an|
+ 1
Example B. For P(x) = -2x5 + 6x3 + 2x2 + x – 1,
then Max{|-2|, |6|, |2|, |1|, |-1|} = 6 and |an| = 2,
hence M = 6/2 + 1 = 4 and all the real roots of P(x)
reside in the interval (–4, 4).
Note: If an = 1, then M = largest Coefficient + 1.
We may use The Theorem of Bounds to find decimal
solutions via a graphing calculator or software.

42. Looking for Real Roots of Real Polynomials
Example C. To find the approximate real roots of
P(x) = -2x5 + 6x3 + 2x2 + x – 1 with a graphing
calculator,

43. Looking for Real Roots of Real Polynomials
Example C. To find the approximate real roots of
P(x) = -2x5 + 6x3 + 2x2 + x – 1 with a graphing
calculator, set the plot range of x to be [–4, 4],
the plot range of y, to be say [20, –20].

44. Looking for Real Roots of Real Polynomials
Example C. To find the approximate real roots of
P(x) = -2x5 + 6x3 + 2x2 + x – 1 with a graphing
calculator, set the plot range of x to be [–4, 4],
the plot range of y, to be say [20, –20].

45. Looking for Real Roots of Real Polynomials
We get three roots.
y = -2x5 + 6x3 + 2x2 + x – 1
Example C. To find the approximate real roots of
P(x) = -2x5 + 6x3 + 2x2 + x – 1 with a graphing
calculator, set the plot range of x to be [–4, 4],
the plot range of y, to be say [20, –20].

46. Looking for Real Roots of Real Polynomials
We get three roots. From the tracer-operation,
their approximate values are x ≈ –1.65, 0.40, and 1.89.
y = -2x5 + 6x3 + 2x2 + x – 1
-1.65 0.40 1.89
Example C. To find the approximate real roots of
P(x) = -2x5 + 6x3 + 2x2 + x – 1 with a graphing
calculator, set the plot range of x to be [–4, 4],
the plot range of y, to be say [20, –20].

47. Looking for Real Roots of Real Polynomials
Polynomials with integer coefficients form an
important class of functions.

48. Looking for Real Roots of Real Polynomials
Polynomials with integer coefficients form an
important class of functions. The next theorem gives
all the possible rational roots of such polynomials.

49. Theorem of Rational Roots:
Looking for Real Roots of Real Polynomials
Polynomials with integer coefficients form an
important class of functions. The next theorem gives
all the possible rational roots of such polynomials.

50. Theorem of Rational Roots:
Let P(x) = Axn + an-1xn-1+… a1x + B be a polynomial
where A, an-1, an-2, ..,a1, B, are all integers.
Looking for Real Roots of Real Polynomials
Polynomials with integer coefficients form an
important class of functions. The next theorem gives
all the possible rational roots of such polynomials.

51. Theorem of Rational Roots:
Let P(x) = Axn + an-1xn-1+… a1x + B be a polynomial
where A, an-1, an-2, ..,a1, B, are all integers.
If x = b/a is rational root of P(x),
then b is a factor of B and a is a factor of A.
Looking for Real Roots of Real Polynomials
Polynomials with integer coefficients form an
important class of functions. The next theorem gives
all the possible rational roots of such polynomials.

52. Theorem of Rational Roots:
Let P(x) = Axn + an-1xn-1+… a1x + B be a polynomial
where A, an-1, an-2, ..,a1, B, are all integers.
If x = b/a is rational root of P(x),
then b is a factor of B and a is a factor of A.
Looking for Real Roots of Real Polynomials
Example D.
a. P(x) = 4x + 6
b. P(x) = 6x2 + 7x + 2
Polynomials with integer coefficients form an
important class of functions. The next theorem gives
all the possible rational roots of such polynomials.

53. Theorem of Rational Roots:
Let P(x) = Axn + an-1xn-1+… a1x + B be a polynomial
where A, an-1, an-2, ..,a1, B, are all integers.
If x = b/a is rational root of P(x),
then b is a factor of B and a is a factor of A.
Looking for Real Roots of Real Polynomials
Example D.
a. P(x) = 4x + 6, it's root is x = –3/2.
b. P(x) = 6x2 + 7x + 2
Polynomials with integer coefficients form an
important class of functions. The next theorem gives
all the possible rational roots of such polynomials.
3 is a factor of 6 and
2 is a factor of 4.

54. Theorem of Rational Roots:
Let P(x) = Axn + an-1xn-1+… a1x + B be a polynomial
where A, an-1, an-2, ..,a1, B, are all integers.
If x = b/a is rational root of P(x),
then b is a factor of B and a is a factor of A.
Looking for Real Roots of Real Polynomials
Example D.
a. P(x) = 4x + 6, it's root is x = –3/2.
b. P(x) = 6x2 + 7x + 2 = (3x + 2)(2x + 1)
It's roots are –2/3, and –1/2.
Polynomials with integer coefficients form an
important class of functions. The next theorem gives
all the possible rational roots of such polynomials.
3 is a factor of 6 and
2 is a factor of 4.

55. Theorem of Rational Roots:
Let P(x) = Axn + an-1xn-1+… a1x + B be a polynomial
where A, an-1, an-2, ..,a1, B, are all integers.
If x = b/a is rational root of P(x),
then b is a factor of B and a is a factor of A.
Looking for Real Roots of Real Polynomials
Example D.
a. P(x) = 4x + 6, it's root is x = –3/2.
b. P(x) = 6x2 + 7x + 2 = (3x + 2)(2x + 1)
It's roots are –2/3, and –1/2.
The numerators of the roots are 2, 1 and are factors
of 2. The denominators 3 and 2 are factors of 6.
Polynomials with integer coefficients form an
important class of functions. The next theorem gives
all the possible rational roots of such polynomials.
3 is a factor of 6 and
2 is a factor of 4.

56. Looking for Real Roots of Real Polynomials
Example E.
a. Let P(x) = 2x3 – 11x2 + 10x + 3, list all the possible
rational roots of P(x).

57. Looking for Real Roots of Real Polynomials
Example E.
a. Let P(x) = 2x3 – 11x2 + 10x + 3, list all the possible
rational roots of P(x).
The factors of 3 are B = {1, 3}.

58. Looking for Real Roots of Real Polynomials
Example E.
a. Let P(x) = 2x3 – 11x2 + 10x + 3, list all the possible
rational roots of P(x).
The factors of 3 are B = {1, 3}.
The factors of 2 are A = {1, 2}

59. Looking for Real Roots of Real Polynomials
Example E.
a. Let P(x) = 2x3 – 11x2 + 10x + 3, list all the possible
rational roots of P(x).
The factors of 3 are B = {1, 3}.
The factors of 2 are A = {1, 2}
The possible roots are fractions of the form ±b/a
where b is from the set B and a is from the set A.

60. Looking for Real Roots of Real Polynomials
Example E.
a. Let P(x) = 2x3 – 11x2 + 10x + 3, list all the possible
rational roots of P(x).
The factors of 3 are B = {1, 3}.
The factors of 2 are A = {1, 2}
The possible roots are fractions of the form ±b/a
where b is from the set B and a is from the set A.
They are {± , ± , ± , ± }.
1
1
3
1
1
2
3
2

61. Looking for Real Roots of Real Polynomials
Example E.
a. Let P(x) = 2x3 – 11x2 + 10x + 3, list all the possible
rational roots of P(x).
The factors of 3 are B = {1, 3}.
The factors of 2 are A = {1, 2}
The possible roots are fractions of the form ±b/a
where b is from the set B and a is from the set A.
They are {± , ± , ± , ± }.
1
1
3
1
1
2
3
2
b. Factor P(x) into real factors completely.

62. Looking for Real Roots of Real Polynomials
Example E.
a. Let P(x) = 2x3 – 11x2 + 10x + 3, list all the possible
rational roots of P(x).
The factors of 3 are B = {1, 3}.
The factors of 2 are A = {1, 2}
The possible roots are fractions of the form ±b/a
where b is from the set B and a is from the set A.
They are {± , ± , ± , ± }.
1
1
3
1
1
2
3
2
b. Factor P(x) into real factors completely.
By trial and error, use synthetic division, we find that
x = 3/2 is a root.

63. Looking for Real Roots of Real Polynomials
Example E.
a. Let P(x) = 2x3 – 11x2 + 10x + 3, list all the possible
rational roots of P(x).
The factors of 3 are B = {1, 3}.
The factors of 2 are A = {1, 2}
The possible roots are fractions of the form ±b/a
where b is from the set B and a is from the set A.
They are {± , ± , ± , ± }.
1
1
3
1
1
2
3
2
b. Factor P(x) into real factors completely.
By trial and error, use synthetic division, we find that
x = 3/2 is a root.
2 –11 10 3
3/2

64. Looking for Real Roots of Real Polynomials
Example E.
a. Let P(x) = 2x3 – 11x2 + 10x + 3, list all the possible
rational roots of P(x).
The factors of 3 are B = {1, 3}.
The factors of 2 are A = {1, 2}
The possible roots are fractions of the form ±b/a
where b is from the set B and a is from the set A.
They are {± , ± , ± , ± }.
1
1
3
1
1
2
3
2
b. Factor P(x) into real factors completely.
By trial and error, use synthetic division, we find that
x = 3/2 is a root.
2 –11 10 3
3/2
2

65. Looking for Real Roots of Real Polynomials
Example E.
a. Let P(x) = 2x3 – 11x2 + 10x + 3, list all the possible
rational roots of P(x).
The factors of 3 are B = {1, 3}.
The factors of 2 are A = {1, 2}
The possible roots are fractions of the form ±b/a
where b is from the set B and a is from the set A.
They are {± , ± , ± , ± }.
1
1
3
1
1
2
3
2
b. Factor P(x) into real factors completely.
By trial and error, use synthetic division, we find that
x = 3/2 is a root.
2 –11 10 3
3/2
2
3
–8
-12
–2
–3
0

66. Looking for Real Roots of Real Polynomials
Hence
2x3 – 11x2 + 10x + 3
= (x – 3/2)(2x2 – 8x – 2)
2 –11 10 3
3/2
2
3
–8
-12
–2
–3
0

67. Looking for Real Roots of Real Polynomials
Hence
2x3 – 11x2 + 10x + 3
= (x – 3/2)(2x2 – 8x – 2)
= (x – 3/2) 2 (x2 – 4x – 1)
2 –11 10 3
3/2
2
3
–8
-12
–2
–3
0

68. Looking for Real Roots of Real Polynomials
Hence
2x3 – 11x2 + 10x + 3
= (x – 3/2)(2x2 – 8x – 2)
= (x – 3/2) 2 (x2 – 4x – 1)
= (2x – 3)(x2 – 4x – 1)
2 –11 10 3
3/2
2
3
–8
-12
–2
–3
0

69. Looking for Real Roots of Real Polynomials
Hence
2x3 – 11x2 + 10x + 3
= (x – 3/2)(2x2 – 8x – 2)
= (x – 3/2) 2 (x2 – 4x – 1)
= (2x – 3)(x2 – 4x – 1)
x2 – 4x – 1 is an irreducible quadratic polynomial,
2 –11 10 3
3/2
2
3
–8
-12
–2
–3
0

70. Looking for Real Roots of Real Polynomials
Hence
2x3 – 11x2 + 10x + 3
= (x – 3/2)(2x2 – 8x – 2)
= (x – 3/2) 2 (x2 – 4x – 1)
= (2x – 3)(x2 – 4x – 1)
x2 – 4x – 1 is an irreducible quadratic polynomial,
by the quadratic formula x = 2 ± 5.
2 –11 10 3
3/2
2
3
–8
-12
–2
–3
0

71. Looking for Real Roots of Real Polynomials
Hence
2x3 – 11x2 + 10x + 3
= (x – 3/2)(2x2 – 8x – 2)
= (x – 3/2) 2 (x2 – 4x – 1)
= (2x – 3)(x2 – 4x – 1)
x2 – 4x – 1 is an irreducible quadratic polynomial,
by the quadratic formula x = 2 ± 5.
Therefore, P(x) factors completely into real factors:
2x3 – 11x2 + 10x + 3
= (2x – 3)(x – (2 + 5))(x – (2 – 5)).
2 –11 10 3
3/2
2
3
–8
-12
–2
–3
0

72. Exercise A. (Descartes' Rule of Signs)
Determine the possible number of positive roots and negative
roots of the following polynomials.
Looking for Real Roots of Real Polynomials
B. (Theorem on the Bounds) Gives an interval where the
roots of the following polynomials must reside.
1. P(x) = x3 + x2 + x + 1 2. P(x) = x3 + x2 + x – 1
3. P(x) = x3 + x2 – x + 1 4. P(x) = x3 + x2 – x – 1
5. P(x) = x3 – x2 – x + 1 6. P(x) = x3 – x2 – x – 1
7. What can we say about the roots of a polynomial with
only all positive even or all negative even degrees of x’s?
8. What can we conclude about the roots of a polynomial
with only odd degrees of x’s?
1. P(x) = x5 + 6x3 + 2x2 – 1 2. P(x) = x4 + 0.01x3 + 0.23x2 – 1/π
3. By the sign-rule, there is at least one positive real root for
P(x) = x4 – 12x3 + 6.8x2 – √101. Graph P(x) using a calculator
over a chosen interval to see if there are more roots.

73. Looking for Real Roots of Real Polynomials
C. (Rational Roots and Factoring Polynomials)
List all the possible rational roots of the following polynomials.
Then find all the rational and irrational roots (all roots are real),
and factor each completely.
1. P(x) = x3 – 2x2 – 5x + 6 2. P(x) = x3 – 3x2 –10x + 6
3. P(x) = 2x3 + 3x2 – 11x – 6 4. P(x) = 3x3 – 4x2 –13x – 6
5. P(x) = –6x3 –13x2 – 4x + 3
6. P(x) = 12x4 – 8x3 – 21x2 + 5x + 6
7. P(x) = 3x4 – x3 – 24x2 – 16x + 8
8. P 𝑥 = 6𝑥4 + 5𝑥3 − 24𝑥2 − 12𝑥 + 16

74. (Answers to the odd problems) Exercise A.
Exercise B.
1. 𝑃(𝑥) has no positive roots and 3 or 1 negative roots.
3. 𝑃(𝑥) has 2 or 0 positive roots and 1 negative root.
5. 𝑃(𝑥) has 2 or 0 positive roots and 1 negative root.
7. It has 0 positive roots and 0 negative roots.
1. 𝑀 =
max{ 1 , 6 , 2 , −1 }
|1|
+ 1 = 7, so the roots reside in (-7,7)
3. 𝑃 𝑥 = 𝑥4– 12𝑥3 + 6.8𝑥2 − 101
The roots are in (-13,13)
Looking for Real Roots of Real Polynomials

75. Exercise C.
1. The possible roots are {±6, ±3, ±2, ±1}
𝑃(𝑥) = (𝑥 − 3)(𝑥 − 1)(𝑥 + 2)
3. The possible roots are {±1, ±
1
2
, ±2, ±3, ±6, ±
3
2
}
𝑃(𝑥) = (𝑥 + 3)(2𝑥 + 1)(𝑥 − 2)
5. The possible roots are {±1, ±
1
2
, ±
1
3
, ±
1
6
, ±3, ±
3
2
}
𝑃 𝑥 = − 2𝑥 + 3 (3𝑥 − 1)(1 + 𝑥)
7. The possible roots are {±1, ±
1
3
, ±2, ±
2
3
, ±4, ±
4
3
, ±8, ±
8
3
}
𝑃 𝑥 = 3𝑥 − 1 𝑥 + 2 (𝑥 + 5 − 1)(𝑥 − 5 − 1)
Looking for Real Roots of Real Polynomials