Answer:
of a polynomial function with integer coefficients.
Rational Zeros Theorem:
If the polynomial ( ) 1
1 1 ... n n P x ax a x ax a n n
− = + ++ − + 0 has integer
coefficients, then every rational zero of P is of the form
p
q
where p is a factor of the constant coefficient 0 a
and q is a factor of the leading coefficient n a
Example 1: List all possible rational zeros given by the Rational Zeros Theorem of
P(x) = 6x
4
+ 7x
3
- 4 (but don’t check to see which actually are zeros) .
Solution:
Step 1: First we find all possible values of p, which are all the factors
of . Thus, p can be ±1, ±2, or ±4. 0 a = 4
Step 2: Next we find all possible values of q, which are all the factors
of 6. Thus, q can be ±1, ±2, ±3, or ±6. n a =
Step 3: Now we find the possible values of p
q by making combinations
of the values we found in Step 1 and Step 2. Thus, p
q will be of
the form factors of 4
factors of 6 . The possible p
q are
12412412412
, , , , , , , , , , ,
11122233366
± ± ± ± ± ± ± ± ± ± ±±
4
6
Example 1 (Continued):
Step 4: Finally, by simplifying the fractions and eliminating duplicates,
we get the following list of possible values for p
q .
1124 1, 2, 4, , , , , 2333
±± ± ± ± ± ± ±
1
6
Now that we know how to find all possible rational zeros of a polynomial, we want to
determine which candidates are actually zeros, and then factor the polynomial. To do this
we will follow the steps listed below.
Finding the Rational Zeros of a Polynomial:
1. Possible Zeros: List all possible rational zeros using the Rational Zeros
Theorem.
2. Divide: Use Synthetic division to evaluate the polynomial at each of the
candidates for rational zeros that you found in Step 1. When the
remainder is 0, note the quotient you have obtained.
3. Repeat: Repeat Steps 1 and 2 for the quotient. Stop when you reach a
quotient that is quadratic or factors easily, and use the quadratic formula
or factor to find the remaining zeros.
Example 2: Find all real zeros of the polynomial P(x) = 2x
4
+ x
3
– 6x
2
– 7x – 2.
Solution:
Step 1: First list all possible rational zeros using the Rational Zeros
Theorem. For the rational number p
q to be a zero, p must be a
factor of a0 = 2 and q must be a factor of an = 2. Thus the
possible rational zeros, p
q , are
1 1, 2, 2
±± ±
Example 2 (Continued):
Step 2: Now we will use synthetic division to evaluate the polynomial at
each of the candidates for rational zeros we found in Step 1.
When we get a remainder of zero, we have found a zero.
Since the remainder is not zero,
12 1 6 7 2
23 31
0
23 3
1 is not a
10 12
zer
o
←
+
−− −
− −
−− −
Since the remainder is zero,
12 1 6 7 2
21
1 is
5 2
21520
a zero
− −−−
−
− −
−
− ←
This also tells us that P factors as
2x
4
+ x
3
– 6x
2
– 7x – 2 = (x + 1)(2x
3
– x
2
– 5x – 2)
Step 3: We now repeat the process on the quotient polynomial
2x
3
– x
2
– 5x – 2. Again using the Rational Zeros Theorem, the
possible rational zeros of this polynomial are
1 1, 2, 2
±± ± .
Since we determined that +1 was not a rational zero in Step 2,
we do not need to test it again, but we should test –1 again.
Since the remainder is zero,
12 1 5 2
232
1 is again a zero
2 3 2 0
− −−
−
−
−
− − ←
Thus, P factors as
2x
4
+ x
3
– 6x
2
– 7x – 2 = (x + 1)(2x
3
– x
2
– 5x – 2)
= (x + 1) (x + 1)(2x
2
– 3x – 2)
= (x + 1)2 (2x
2
– 3x – 2)
Example 2 (Continued):
Step 4: At this point the quotient polynomial, 2x
2
– 3x – 2, is quadratic.
This factors easily into (x – 2)(2x + 1), which tells us we have
zeros at x = 2 and 1
2
x = − , and that P factors as
2x
4
+ x
3
– 6x
2
– 7x – 2 = (x + 1)(2x
3
– x
2
– 5x – 2)
= (x + 1) (x + 1)(2x
2
– 3x – 2)
= (x + 1)2 (2x
2
– 3x – 2)
= (x + 1)2 (x – 2)(2x + 1)
Step 5: Thus the zeros of P(x) = 2x
4
+ x
3
– 6x
2
– 7x – 2 are x = –1, x = 2,
and 1
2
x = − .
Descartes’ Rule of Signs and Upper and Lower Bounds for Roots:
In many cases, we will have a lengthy list of possible rational zeros of a polynomial. A
theorem that is helpful in eliminating candidates is Descartes’ Rule of Signs.
In the theorem, variation in sign is a change from positive to negative, or negative to
positive in successive terms of the polynomial. Missing terms (those with 0 coefficients)
are counted as no change in sign and can be ignored. For example,
has two variations in sign.
Descartes’ Rule of Signs: Let P be a polynomial with real coefficients
1. The number of positive real zeros of P(x) is either equal to the number
of variations in sign in P(x) or is less than that by an even whole
number.
2. The number of negative real zeros of P(x) is either equal to the number
of variations in sign in P(–x) or is less than that by an even whole
number.
Example 3: Use Descartes’ Rule of Signs to determine how many positive and how
many negative real zeros P(x) = 6x
3
+ 17x
2
– 31x – 12 can have. Then
determine the possible total number of real zeros.
Solution:
Step 1: First we will count the number of variations in sign of
( ) . 3 2 Px x x x =+ −− 6 17 31 12
Explanation: