### Section 2.5 Part 2 Other tests for Zeros

```Section 2.5 Part Two
Other Tests for Zeros
Descartes’s Rule of Signs
Upper and Lower Bounds
Descartes’s Rule of Signs
• Let f(x) be a polynomial with real coefficients and
ao ≠ 0.
f ( x )  a n x  a n 1 x
n
n 1
 ...  a 2 x  a1 x  a 0
2
• The number of positive real zeros of f is either
equal to the number of variations in sign of f(x) or
less than that number by an even integer.
• The number of negative real zeros of f is either
equal to the number of variations in the sign of f(-x)
of less than that number by an even integer.
Apply Descarte’s Rule of Signs
• Consider f(x) = x2 + 8x + 15
• Since there are no variations in sign of f(x)
there are no positive roots.
• Since f(-x) = (-x)2 + 8(-x) + 15
= x2 – 8x + 15
has two variations, f(x) may have two or zero
negative roots.
x = {-3. -5}
Variation in Sign
• A variation in sign means that when the
polynomial is written in standard form that one
term has a different sign than the next.
• T(x) = x5 + 9x4 + 19x3 – 21x2 – 92x – 60
• For Variations of T(x) just look at the signs
T(x) has one variation in signs
• For Variations of T(-x) you must change the sign
of terms with odd numbered exponents
• T(-x) = -x5 + 9x4 – 19x3 – 21x2 + 92x – 60
T(-x) has four variation in signs
Find the Zeros of T(x)
• T(x) = x5 + 9x4 + 19x3 – 21x2 – 92x – 60
• 60  + 1, + 2 , + 3, + 4, + 5 , + 6 , + 10 ,
+ 12 , + 15 , + 20, + 30, + 60
• Since there is only one sign variation for
T(x) there is at most, one positive root
• SYN Program
Find the Zeros of T(x)
T(x) = x5 + 9x4 + 19x3 – 21x2 – 92x – 60
T(x) = (x – 2)(x4 + 11x3 + 41x2 + 61x + 30)
+2
1
1
9
19
-21 -92 -60
2
22
82 122
11
41
61
30
60
0
Now that we have found the positive root we
know that any other real roots must be negative.
Find the Zeros of T(x)
T(x) = x5 + 9x4 + 19x3 – 21x2 – 92x – 60
T(x) = (x – 2)(x4 + 11x3 + 41x2 + 61x + 30)
• 30  -1, -2 , -3, -4, -5 , -6 , -10 ,-15, - 30
T(x) = (x – 2)(x + 1)(x3 + 10x2 + 31x + 30)
-1
1
1
11 41
61
-1 -10
-31 -30
10
31
30
30
0
Find the Zeros of T(x)
T(x) = x5 + 9x4 + 19x3 – 21x2 – 92x – 60
T(x) = (x – 2)(x4 + 11x3 + 41x2 + 61x + 30)
T(x) = (x – 2)(x + 1)(x3 + 10x2 + 31x + 30)
• 30  -1, -2 , -3, -4, -5 , -6 , -10 ,-15, - 30
T(x) = (x – 2)(x + 1)(x + 2)(x2 + 8x + 15)
-2
1
1
10 31
30
-2 -16
-30
8
15
0
Find the Zeros of T(x)
T(x) = x5 + 9x4 + 19x3 – 21x2 – 92x – 60
T(x) = (x – 2)(x4 + 11x3 + 41x2 + 61x + 30)
T(x) = (x – 2)(x + 1)(x3 + 10x2 + 31x + 30)
T(x) = (x – 2)(x + 1)(x + 2)(x2 + 8x + 15)
T(x) = (x – 2)(x + 1)(x + 2)(x + 3)(x + 5)
x = {-5, -3, -2, -1, 2}
Homework 2.5
•
•
•
•
•
Zeros of Polynomial Functions
page 160
1 - 31 odd,
37 - 85 odd,
91 - 94 all
#5 Find all of the zeros of the
function.
F(x) = (x + 6)(x + i)(x – i)
x = {-6, -i, i}
#10 Use the rational zero test to list all of the
possible rational zeros of f. Verify that the
zeros of f shown are contained in the list.
f(x) = 4x5 – 8x4 – 5x3 + 10x2 + x – 2
4
Factors of 2
Factors of 4
2

1  2
1  2  4
1  2 
-2
-4
-6
1
2

1
4
1 1
x  {  1,
, , 1, 2}
2 2
#15 Find all of the real zeros of the
function.
h(t) = t3 + 12t2 + 21t + 10
Factors of 10
Factors of 1

 1  2  5  10
1
Caution graph may be
6
With a cubic we expect
another turn down to the left
4
2
x  {  10 ,  1,  1}
-5
5
-2
#25 f(x) =
x3
Factors of 4

Factors of 1
+
x2
– 4x – 4
1  2  4
1
(a)
(b)
(c)
List the possible rational zeros of f,
sketch the graph of f so that some
possibilities can be eliminated
determine all the real zeros
1,  1, 2 ,  2 , 4 ,  4
4
Graph eliminates -4, 1, 4
f(x) = x3 + x2 – 4x – 4
2
f(x) = x2 (x +1)– 4(x + 1)
-5
5
f(x) = (x + 1)(x2– 4)
-2
f(x) = (x + 1)(x +2)(x – 2)
-4
-6
x  {  2 ,  1, 2}
#42. Find the polynomial function with
integer coefficients that has the given zeros.
 5, 5, 1 
Since imaginary solutions always
appear in conjugate pairs we
know to include
3i
1
f ( x )  ( x  5 )( x  5 )( x  1 
3 i )( x  1 
f ( x )  ( x  25 )( x  2 x  4 )
2
2
f ( x )  x  2 x  21 x  50 x  100
4
3
2
3i
3i )
```