2. Example 1 ±1 ± 2 Disregard the negative sign Factors of the constant:
±1 ± 2
Disregard the negative sign
Factors of the constant:
± 1 ± 6
± 2 ± 3
Factors of the leading coefficient:
According to the “Rational Root Theorem,” the possible rational roots are all combinations of the values of the:
This reduces the number of possible rational roots from an infinite number down to 16 possibilities. Can some of these possibilities be eliminated?
We are down to 12 possible roots, now evaluate each to see where the zeros are.

3. Example 1 (cont’d) Possible rational solutions for
The polynomial has a degree of three, and there are three zero values, so all the roots are real #’s. These roots are:
1/2, -2, & -1/3

4. Example 2
Since the leading coefficient is 1, any possible rational roots will be factors of the constant (5).
The factors of 5 are: 5
-1 -5
-5 is the only rational root; the other two are either irrational or imaginary.
Test these factors:
Use long or synthetic division to find the imaginary roots: 1
The remainder better be zero or – 5 is not one of the roots!
- 5
- 15
- 5 3 1
The depressed polynomial is:
Which can be solved by factoring, completing the square or quadratic formula.
The 3 roots are:

5. Example 3 – Descartes’ Rule of Signs
The rule: The number of positive real zeros is the number of sign changes (in this case 4) or any even number less that the number of sign changes.
There will be either 4, 2, or 0 positive real zeros.
The rule: The number of negative real zeros is the number of sign changes (in this case 1) or less than this number by an even number.
There will be 1 negative real zero.

6. Example 3 (cont’d) To determine the rational roots:
There will be 4, 2, or zero positive real roots & 1 negative real root.
Example 3 (cont’d)
To determine the rational roots:
Factors of the constant:
±1 ± 3
Factors of the leading coefficient:
±1 ± 2
Possible rational zeros:
Any duplicates to throw out?
Test them:

7. There will be 4, 2, or zero positive real roots & 1 negative real root.
Example 3 (cont’d)
Did I really have to test these? There could not be a third positive or a second negative, so it was a waste of time.
There are 2 positive real zeros: 1 & ½
There is 1 negative real zero: -3
There are still two remaining unknown roots, so they must be imaginary.
The question was limited to only the rational zeros.

8. Example 4 Define the variables: Width = w Height = ½ w + 1
Length = 324 w + 32
The formula for volume of a rectangular prism is:
Multiply it out
Rewrite equal to zero
Everything divides by 2 to make it smaller numbers to work with. No
Yes
Sign changes?
There is only one sign change, so there is only one positive real root. Since dimensions are always positive, we don’t need to worry about the negative real roots.

9. Example 4 Define the variables: Width = w Height = ½ w + 1
Length = 324 w + 32
Easiest way to solve by hand: Select values for w & test in 81w3+170w2+16w until the total is 31,115,520. w
81w3+170w2+16w
V(w) 20
716,320
Way too low
Compare V(w) to 31,115,520
100
82,701,600
Way too high 60
18,108,960
Way too low 80
42,561,280
Still too high
Width = w = 72 ft 70
28,617,120
Still too low
Height = ½ w + 1= 37 ft 71
29,848,897
Still too low
Length = 324 w = 23,360 ft (over 4 mi) 72
31,115,520
Just right!

10. HW: Page 233