Transcript Section 5.2

The factors of the constant.
The factors of the leading coefficient.
The factors of the 12.
The factors of the 2.
=
1, 2, 3, 4, 6, 12
1, 2
1 3
+
{1,
2,
3,
4,
6,
12,
, }
=
2 2
3
1 1
3 2, 3, 4, 6, 12
 12,  6,  4,  3, 2,  ,  1,  , , 1, ,
2
2
2 2
{
}
Example
x 2  5x  6
x  1 x3  6 x 2  11x  6
x 3  1x 2
x3  6 x 2  11x  6  x 2  5x  6 x  1
5 x 2  11x
 x  2x  3x  1
5x 2  5x

6x  6
6x  6
0

Synthetic Division is used to quickly check, if possible rational zeros are factors. Test x = -1.
x
3

 6 x 2  11x  6
x  1 Synthetic division can only be done if the degree of the divisor is 1.
x+1=0 1
x=–1
6
11
6
–1
–5
–6
–1
5
6
0
1
Move the first number down to the
bottom row. Multiply the potential zero
to the bottom row number and move the
product up to the next column to
combine. Repeat the process until you
have the remainder at the bottom of the
last column.
x
2

 5x  6 x  1
x  2x  3x 1
+ + + + + + +
+ – + – + – +
Positive signs
Alternating signs
Zero can be used for any needed sign!
ONLY TEST
ONLY TEST
POSITIVE
NEGATIVE
RATIONAL
RATIONAL
ZEROS!
ZEROS!
Upper bound
Lower bound
2
7
-7
-12
-4
-6
26
-2 2
3
-13
14
-2 is not a lower bound the
signs do not alternate.
2
-4
2
7
-7
-12
-8
4
12
-1
-3
0
-4 is not a lower bound, but
it is a zero.
2
7
6
-7
-12
39
96
3
2
13 32
84
3 is an upper bound the
signs are all positive.
x  42 x 2  x  3  0
x  42 x  3x  1  0
3
x  4 x 
2
x  1
3
2
Find the zeros of f x   2 x  11x  7 x  6
zeros are x-intercepts,
soooooo lets graph it!
Using the possible
rational zero theorem,
the smallest number is
-6 and the largest is 6.
Set your window or
Zoom 6.
Count tick marks to
make logical guesses!
Hit Trace and type in
your guess.
.
Zeros are x = -6 , x = 1, x = - ½
Find the zeros of f x   x 5  3x 4  10 x 3  22 x 2  7 x  3
Zeros are x = 3 , x = -1 , -1
WAIT!! Take another LOOK!
It doesn’t cross the x-axis!
What does this mean?
-1 has even multiplicity.
Probably 2. Why?
The graph is touching. or
crossing the x-axis 4 times
and with a 5th degree we
need at most 5 zeros.
Now we will use the existing 3
rational zeros to find the other 2
zeros that are not rational numbers.
Find the zeros of f x   x 5  3x 4  10 x 3  22 x 2  7 x  3
Zeros are x = 3 , x = -1, -1
Start by factoring
out the two -1 zeros.
1
-1
1
3
-10
-22
-7
3
-1
-2
12
10
-3
2
-12
-10
3
0
Remainder
-1
1
x  4 x   2  1  4
x  22  5
2
2
x2 5
x  2  5
-1
13
-3
1
-13
3
0
Remainder
Now the 3..
x 2  4 x  __  1  __
-1
3
1
3
12
-3
4
-1
0
x2  4x 1  0
Remainder
Complete the square
x  2  5
x  3,1,1
Means
ZERO
Find the zeros of f x  x 5  3x 4  5x 3  23x 2  24 x  8
Zeros are x = 1 , 1 , 1
WAIT!! Take another LOOK!
It crosses the x-axis like
what type of curve?
A Cubic curve. 1 has odd
multiplicity. Probably 3. Why?
The graph is crossing the
x-axis in two other
locations and with a 5th
degree we need at most 5
zeros.
Now we will use the existing 3
rational zeros to find the other 2
zeros that are not rational numbers.
Find the zeros of f x  x 5  3x 4  5x 3  23x 2  24 x  8
1
Zeros are x = 1 , 1, 1
Start by factoring
out all the 1’s.
1
1
-3
-5
23
-24
8
1
-2
-7
16
-8
-2
-7
16
-8
0
Remainder
x 8  0
2
x2  8
x 8
x  2 2
x  2 2
x  1,1,1
1
1
1
-1
-8
8
-1
-8
8
0
Remainder
.
1
1
1
0
-8
0
-8
0
x2  8  0
Solve for x.
Means
ZERO
Remainder
c
= f(c)
1
-2
1
When an x-term is missing, put in a
zero to hold the place of the term.
0
-12
6
-2
4
16
-2
-8
22 = f(-2)
4
1
3
1
-2
4
-25
0
-4
3
3
21
-12
-36
1
7
-4
-12
-40 = f(3)