Download Sect 5.7 Part b - Synthetic Division

A– 1
Sect 5.7 Part b - Synthetic Division
Objective a: Understanding and Using Synthetic Division.
When dividing a polynomial by binomial in the form of x – c, we can use a
process called synthetic division. The advantage of synthetic division is
that it is easier to perform than long division. The disadvantages are that
can only be used when the divisor is in the form x – c (binomial of degree
one, coefficient of the variable term is one). To see where synthetic
division comes from, let’s look at a long division.
Perform the indicated operation:
Ex. 1
(x2 + 11x – 19) ÷ (x – 4)
Solution:
x2
x
= x, subtract x(x – 4) which is
the same as adding – x2 + 4x.
Bring down the – 19.
15x
x
= 15,
x + 15
x – 4 x2 + 11x – 19
– x2 + 4x
subtract 15(x – 4) or add – 15x + 60.
Put the remainder over the divisor.
So, the answer is x + 15 +
41
.
x−4
15x – 19
– 15x + 60
41
The important information in the problem are the coefficients. So, let’s drop
the variables and only list the variables.
x + 15
x – 4 x + 11x – 19
– x2 + 4x
15x – 19
– 15x + 60
41
2
1 + 15
1 – 4 1 + 11 – 19
–1 + 4
15 – 19
– 15 + 60
41
– (– 4•1)
– (– 4•15)
The + 4 came from subtracting a negative 4 times 1. The + 60 came from
subtracting a negative 4 times 15. But subtracting a negative is the same
as adding a positive, so to make this easier, we will change the sign of the
– 4 in the divisor. Also, we do not need to keep track of the – 1 and – 15 so
we will drop them. Lastly, we will drop the plus signs.
A– 2
Drop the numbers in bold, change the sign of – 4 to 4
1 + 15
1 – 4 1 + 11 – 19
–1 + 4
15 – 19
– 15 + 60
41
4 1
1 15
11 – 19
4
15 – 19
60
41
(4•1)
(4•15)
So, we take 4 times 1, add to 11 to get 15. Then, we take 4 times 15 and
add – 19. The answer is the remainder and the 1 and 15 are the
coefficients. To make this easier to read, we will reorganize the algorithm.
Delete the coefficients of the divisor, and make the 1, 15, and 41 along the
diagonal the bottom row.
4 1
1 15
11 – 19
4
15 – 19
60
41
4
1
1
11
4
15
– 19
60
41
Now, let’s run through the algorithm: (x2 + 11x – 19) ÷ (x – 4)
Since the divisor is in the form of x – c, c is 4. We write 4 in the upper left.
Next, list the coefficients of the dividend in the upper right. Bring down the
first coefficient (1), multiply it by 4 and add the result to the second
coefficient (11) to get 15. Finally, multiply 15 by 4 and add the result to the
third coefficient (– 19) to get 41. The last number, 41, is the divisor, the
other numbers are the coefficients of the quotient.
4
1
1
4• 1
11
4
15
– 19
60
4•15 41
Note the degree of the quotient will always be one less than the degree of
the dividend whenever we are using synthetic division. Since the degree of
the dividend was 2, the degree of the quotient will be one. It is also
important that the polynomials be listed in order of descending powers
before starting the synthetic division. Finally, if there are terms missing, we
will need to use zeros as place holders in the synthetic division.
A– 3
Use synthetic division to divide:
Ex. 2
(x3 – 7x2 – 13x + 3) ÷ (x + 2)
Solution:
Since x + 2 = x – (– 2), c = – 2.
–2
1 – 7 – 13
3
–2
18 – 10
Bring down the 1. – 2•1 = – 2.
List it under – 7 and add to get
1 –9
5 –7
– 9. – 2(– 9) = 18. List it under – 13
and add to get 5. – 2•5 = – 10. List it under 3 and add to get – 7.
1 is the coefficient of the x2, – 9 the coefficient of x, five is the
constant term and – 7 is the remainder. So our answer is
x2 – 9x + 5 –
Ex. 3
(6x3 – 3x2 – 45x – 9) ÷ (x – 3)
Solution:
c= 3
3
6 – 3 – 45
Bring down the 6. 3•6 = 18.
18
45
List it under – 3 and add to get
6 15
0
15. 3(15) = 45. List it under – 45
and add to get 0. 3•0 = 0. List it under – 9 and add to get – 9.
6 is the coefficient of the x2, 15 the coefficient of x, zero is the
constant term and – 9 is the remainder. So our answer is
6x2 + 15x –
Ex. 4
7
.
x+2
–9
0
–9
9
.
x−3
(x3 – 64) ÷ (x – 4)
Solution:
Since the dividend has no x2 and x terms, we will need to use zeros
as place holders in the synthetic division.
c=4
4
1
0
0 – 64
4
16
64
Bring down the 1. 4•1 = 4.
List it under the first 0 and add to get
1
4
16
0
4. 4(4) = 16. List it under the second 0
and add to get 16. 4•16 = 64. List it under – 64 and add to get 0.
1 is the coefficient of the x2, 4 the coefficient of x, sixteen is the
constant term and 0 is the remainder. So our answer is
x2 + 4x + 16 .
A– 4
1
)
2
(8r4 – 2 + 12r + 10r3 – 5r2) ÷ (r +
Ex. 5
Solution:
First, write the polynomials in order of descending powers:
(8r4 + 10r3 – 5r2 + 12r – 2) ÷ (r +
Since r +
–
1
2
1
2
= r – (–
1
),
2
c=–
1
)
2
1
.
2
8
10
–5
12
–2
8
–4
6
–3
–8
4 –8
16 – 10
10
So, the answer is 8r3 + 6r2 – 8r + 16 –
r+
= 8r3 + 6r2 – 8r + 16 –
Ex. 6
1
2
20
.
2r +1
(3x3 – 2x2 + 6x + 5) ÷ (x2 + 1)
Solution:
Since the degree of the divisor is not 1, we cannot use synthetic
division on this problem. We have to use long division.
3x3
2
x
= 3x, subtract 3x(x2 + 1) or add – 3x3 – 3x, being careful to
align like terms. Bring down the 5.
2
−2x2
x
2
= – 2, subtract – 2(x2 + 1)
or add 2x + 2, being careful to line up like terms.
Since the degree of the divisor is higher than what is left, we stop
and put the remainder over the divisor.
3x – 2 +
x2 + 1
3x3 – 2x2 + 6x + 5
– 3x3
– 3x
2
– 2x + 3x + 5
2x2
2
3x + 7
3x + 7
x2 + 1
A– 5
a) Divide f(x) by the x + 3 and then b) find f(– 3).
Ex. 7
f(x) = 6x4 + 15x3 + 28x + 5
Solution:
a) Here, we can use synthetic division. Since x + 3 = x – (– 3),
c = – 3. Also, the dividend has no x2, we will need to a zero as a
place holder:
–3
6
15
– 18
6 –3
0
28
9 – 27
9
1
5
–3
2
So, the answer is 6x3 – 3x2 + 9x + 1 +
2
.
x+ 3
b) f(– 3) = 6(– 3)4 + 15(– 3)3 + 28(– 3) + 5
= 6(81) + 15(– 27) + 28(– 3) + 6 = 486 – 405 – 84 + 5 = 2
Notice that the answer is the same as the remainder. To see why
f(– 3) is equal to remainder, recall that the dividend is equal to the
quotient times the divisor plus the remainder. So,
f(x) = (6x3 – 3x2 + 9x + 1)(x + 3) + 2 = 6x4 + 15x3 + 28x + 5
Thus, f(– 3) = (stuff)(– 3 + 3) + 2 = (stuff)(0) + 2 = 0 + 2 = 2. This is
known as the remainder theorem.
The Remainder Theorem
The remainder obtained by dividing the polynomial P(x) by x – r is P(r).
Find P(2) by synthetic division:
Ex. 8
P(x) = x4 + 7x3 + 11x2 – 7x – 12
Solution:
Here, c = 2, so
2
1
1
7
2
9
11
18
29
–7
58
51
– 12
102
90
Since the remainder is 90, then P(2) = 90.