#### Transcript Slide 1

```10TH
EDITION
COLLEGE
ALGEBRA
LIAL
HORNSBY
SCHNEIDER
3.2 - 1
3.2 Synthetic Division
Synthetic Division
Evaluating Polynomial Functions Using
the Remainder Theorem
Testing Potential Zeros
3.2 - 2
Division Algorithm
Let (x) and g(x) be polynomials with g(x) of
lower degree than (x) and g(x) of degree one
or more. There exists unique polynomials q(x)
and r(x) such that
f  x   g  x  q  x   r  x ,
where either r(x) = 0 or the degree of r(x) is
less than the degree of g(x).
3.2 - 3
Synthetic Division
Synthetic division is a shortcut method of
performing long division with polynomials.
It is used only when a polynomial is divided
by a first-degree binomial of the form x – k,
where the coefficient of x is 1.
3.2 - 4
Synthetic Division
3x  10x  40
3
2
x  4 3 x  2 x  0 x  150
3 x 3  12 x 2
10 x 2  0 x
2
10 x  40 x
40x  150
40x  160
10
2
3.2 - 5
Synthetic Division
3 10 40
Here the division
process is simplified 4 3  2  0  150
by omitting the
3  12
variables and writing
10  0
only coefficients, with
10  40
0 used to represent
the coefficient of any
40  150
missing terms.
40  160
10
3.2 - 6
Synthetic Division
The numbers
in color that are
repetitions of
the numbers
directly above
them can be
omitted as
shown here.
3
4 3
10 40
2
0  150
 12
10  0
 40
40  150
160
10
3.2 - 7
Synthetic Division
4 3
2
0  150
 12  40 160
40
10
3 10
The entire problem can now be condensed
vertically, and the top row is obtained by
subtracting –12, –40, and –160 from the
corresponding terms above them.
3.2 - 8
Synthetic Division
inverse
Quotient
2
0  150
12 40
160
40
10
3 10
43
10
3 x  10 x  40 
x4
2
Signs
changed
Remainder
With synthetic division it is helpful to change the sign
of the divisor, so the –4 at the left is changed to 4,
which also changes the sign of the numbers in the
second row. To compensate for this change,
3.2 - 9
Caution To avoid errors, use 0 as
the coefficient for any missing terms,
including a missing constant, when
setting up the division.
3.2 - 10
USING SYNTHETIC DIVISION
Example 1
Use synthetic division to divide
5 x  6 x  28 x  2
.
x2
Solution Express x + 2 in the form x – k by
writing it as x – (–2). Use this and the
coefficients of the polynomial to obtain
to –2
2 5
3
2
6
 28
 2.
Coefficients
3.2 - 11
Example 1
USING SYNTHETIC DIVISION
Use synthetic division to divide
5 x  6 x  28 x  2
.
x2
Solution Bring down the 5, and multiply:
–2(5) = –10
2 5  6  28  2
10
5
3
2
3.2 - 12
Example 1
USING SYNTHETIC DIVISION
Use synthetic division to divide
5 x  6 x  28 x  2
.
x2
Solution Add –6 and –10 to obtain –16.
Multiply –2(–16) = 32.
3
2 5
2
 6  28
10
32
5 16
2
3.2 - 13
Example 1
USING SYNTHETIC DIVISION
Use synthetic division to divide
5 x  6 x  28 x  2
.
x2
Solution Add –28 and 32 to obtain 4.
Finally, –2(4) = – 8.
3
2 5
2
 6  28
10
32
5 16
4
2
8
columns.
Watch your
signs.
3.2 - 14
Example 1
USING SYNTHETIC DIVISION
Use synthetic division to divide
5 x  6 x  28 x  2
.
x2
Solution Add –2 and –8 to obtain –10.
3
2
2 5
 6  28  2
10
32 8
5 16
4 10
Remainder
Quotient
3.2 - 15
Example 1
USING SYNTHETIC DIVISION
Use synthetic division to divide
5 x  6 x  28 x  2
.
x2
Since the divisor x – k has degree 1, the
degree of the quotient will always be written
one less than the degree of the polynomial to
be divided. Thus,
3
2
5 x  6 x  28 x  2
10
2
 5 x  16 x  4 
.
x2
x2
Remember to
3
2
divisor
.
3.2 - 16
Special Case of the Division
Algorithm
For any polynomial (x) and any
complex number k, there exists a
unique polynomial q(x) and number r
such that
f ( x )  ( x  k )q( x )  r .
3.2 - 17
For Example
In the synthetic division in Example 1,
5 x 3  6 x 2  28 x  2  ( x  2)(5 x 2  16 x  4)  ( 10).
f (x)
 (x  k )
q( x )

r
Here g(x) is the first-degree polynomial x – k.
3.2 - 18
Remainder Theorem
If the polynomial (x) is divided by
x – k, the remainder is equal to (k).
3.2 - 19
Remainder Theorem
In Example 1, when (x) = 5x3 – 6x2 – 28x – 2
was divided by x + 2 or x –(–2), the
remainder was –10. Substituting –2 for x in
(x) gives
f ( 2)  5( 2)3  6( 2)2  28( 2)  2
 40  24  56  2
 10
Use parentheses
around substituted
values to avoid errors.
3.2 - 20
Remainder Theorem
A simpler way to find the value of a polynomial
is often by using synthetic division. By the
remainder theorem, instead of replacing x by –2
to find (–2), divide (x) by x + 2 using synthetic
division as in Example 1. Then (–2) is the
remainder, –10.
2 5  6  28  2
10
32 8
 (–2)
5 16
4 10
3.2 - 21
Example 2
APPLYING THE REMAINDER
THEOREM
Let (x) = –x4 + 3x2 – 4x – 5. Use the
remainder theorem to find (–3).
Solution Use synthetic division with k = –3.
3 1 0 3  4  5
3  9 18  42
1 3  6 14  47
Remainder
By this result, (–3) = –47.
3.2 - 22
Testing Potential Zeros
A zero of a polynomial function  is a number k
such that (k) = 0. The real number zeros are
the x-intercepts of the graph of the function.
The remainder theorem gives a quick way to
decide if a number k is a zero of the polynomial
function defined by (x). Use synthetic division to
find (k); if the remainder is 0, then (k) = 0 and k
is a zero of (x). A zero of (x) is called a root or
solution of the equation (x) = 0.
3.2 - 23
Example 3
DECIDING WHETHER A NUMBER
IS A ZERO
Decide whether the given number k is a zero
of (x).
3
2
f
(
x
)

x

4
x
 9 x  6; k  1
a.
Solution
Proposed
zero
11 4 9 6
1 3
6
1 3 6
0
f ( x )  x 3  4x 2  9x  6
Remainder
Since the remainder is 0, (1) = 0, and 1 is a zero of the
polynomial function defined by (x) = x3 – 4x2 + 9x – 6. An
x-intercept of the graph (x) is 1, so the graph includes the
point (1, 0).
3.2 - 24
Example 3
DECIDING WHETHER A NUMBER
IS A ZERO
Decide whether the given number k is a zero
of (x).
4
2
f
(
x
)

x

x
 3 x  1; k   4
b.
Solution Remember to use 0 as coefficient for
the missing x3-term in the synthetic division.
Proposed
zero
4 1 0
1 3
1
4 16  68 284
1  4 17  71 285
Remainder
The remainder is not 0, so – 4 is not a zero of
(x) = x4 +x2 – 3x + 1. In fact, (– 4) = 285, indicating that
(– 4, 285) is on the graph of (x).
3.2 - 25
Example 3
DECIDING WHETHER A NUMBER
IS A ZERO
Decide whether the given number k is a zero
of (x).
4
3
2
c. f ( x )  x  2 x  4 x  2 x  5; k  1  2i
Solution Use synthetic division and operations
with complex numbers to determine whether 1
+ 2i is a zero of (x) = x4 – 2x3 + 4x2 + 2x – 5.
1  2i 1  2
4
1  2i  5
(1  2i )( 1  2i )
 1  4 i 2
 5
1  1  2i  1
2
5
 1  2i
5
1  2i
0
Remainder
3.2 - 26
Example 3
DECIDING WHETHER A NUMBER
IS A ZERO
Decide whether the given number k is a zero
of (x).
4
3
2
c. f ( x )  x  2 x  4 x  2 x  5; k  1  2i
Since the remainder is 0, 1 + 2i is a zero of the given
polynomial function. Notice that 1 + 2i is not a real
number zero. Therefore, it would not appear as an
x-intercept in the graph of (x).
1  2i 1  2
4
1  2i  5
(1  2i )( 1  2i )
 1  4 i 2
 5
1  1  2i  1
2
5
 1  2i
5
1  2i
0
Remainder
3.2 - 27
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