Download Sect 3.2 – Synthetic Division

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Sect 3.2 – Synthetic Division
Objective 1:
Division Algorithm
Recall that when dividing two numbers, we can check our answer by the
whole number (quotient) times the divisor plus the remainder. This should
be equal to the original number we were dividing into (dividend).
Division Problem:
217 R 13
36 7825
)
Ans: 217
Check:
7825 = 36•217 + 13
Dividend = Divisor•Quotient + Remainder
13
36
Remainder
Divisor
Quotient
€
We can extend this idea to polynomials:
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Division Problem:
Check:
€
5x – 4 R 11
5x2 + 6x + 3 = (x + 2)•(5x – 4) + 11
)
x + 2 5x 2 + 6x + 3
Ans: 5x – 4 +
€
Quotient +
11
x+2
Remainder
Divisor
Dividend = Divisor•Quotient + Remainder
P(x) = D(x)•Q(x) + R(x)
We call the "check" the Division Algorithm since it relates the relationship
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between division and multiplication.
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Division Algorithm
Let P(x) (Dividend) and D(x) (Divisor) be two polynomials such that degree
of D(x) > 0 and the degree of D(x) is less than P(x). Then, there exists
unique polynomials Q(x) (Quotient) and R(x) (Remainder) such that:
P(x) = D(x)•Q(x) + R(x)
where R(x) = 0 or the degree of R(x) is less than the degree of D(x).
Objective 2:
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 advantages 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
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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.
15x
Bring down the – 19.
= 15,
x + 15
x – 4 x2 + 11x – 19
– x2 + 4x
x
subtract 15(x – 4) or add – 15x + 60.
Put the remainder over the divisor.
41
So, the answer is x + 15 +
.
15x – 19
– 15x + 60
41
x−4
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.
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)
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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.
1 15
4 1 11 – 19
4
1
11 – 19
4
4
60
15 – 19
1
15
41
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
11
– 19
4
60
1
15 4•15 41
4•1
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 placeholders in the synthetic division.
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
Bring down the 1. – 2•1 = – 2.
–2
18 – 10
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, we get x2 – 9x + 5 –
€
7
.
x+2
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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
9
6x2 + 15x –
.
–9
0
–9
x−3
Ex. 4
(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
Bring down the 1. 4•1 = 4.
4
16
64
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 .
(8r4 – 2 + 12r + 10r3 – 5r2) ÷ (r +
Ex. 5
1
)
2
Solution:
First, write the polynomials in order of descending powers:
1
(8r4 + 10r3 – 5r2 + 12r – 2) ÷ (r€+ )
Since r +
–
€
1
2
1
2
= r – (–
8
€ 8
1
),
2
c=–
10
–5
€
–4 –3
€ –8
6
1
.
2
2
12
4 –8
16 – 10
€
So, (8r4 + 10r3 – 5r2 + 12r – 2) ÷ (r +
€
–2
1
)
2
= 8r3 + 6r2 – 8r + 16 –
€
10
r+
1
2
.
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If we were to multiply both sides of the result from the above example by
(r +
1
),
2
we would get:
8r4 + 10r3 – 5r2 + 12r – 2 = (r +
1
)(8r3
2
+ 6r2 – 8r + 16) – 10.
P(x) = (x – c)•Q(x) + r
€
This illustrates the special case of the Division Algorithm:
€ Algorithm
Special Case of the Division
Let P(x) be a polynomial and c be any complex number. Then, there exists
unique polynomial Q(x) and a number r such that:
P(x) = (x – c)•Q(x) + r
It is important to note that one needs to be dividing by a linear factor for the
Special Case of the Division Algorithm to work and to be able to use
synthetic division.
Ex. 6
(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
x2
= 3x, subtract 3x(x2 + 1) or add – 3x3 – 3x, being careful to
align like terms. Bring down the 5.
2
€
−2x2
2
x
= – 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+7
3x – 2 + 2
2
x +1
Objective 3:
3
2
x +1
3x – 2x + 6x + 5
– 3x3
– 3x
2
– 2x + 3x + 5
2x2 €
2
3x + 7
Using the Remainder Theorem to Evaluate Polynomials.
99
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
0
28
5
– 18
9 – 27 – 3
6 –3
9
1
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) + 5 = 486 – 405 – 84 + 5 = 2
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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 – c is P(c).
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.
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Objective 4:
Using the Remainder Theorem to Test Potential Zeros.
A zero of a polynomial P(x) is a number c such that P(c) = 0. If c is a real
number, then (c, 0) is an x-intercept of P(x). We can use the remainder
theorem to quick determine if c is a zero of P(x). If the remainder of
P(x) ÷ (x – c) is zero, then c is a zero and we also have the quotient of P(x)
divided by (x – c).
Determine whether the given number k is a zero of f(x).
Ex. 9
f(x) = 8x3 + 14x2 – 19x – 30; k = – 2
Solution:
Using synthetic division, we get:
– 2
8
14
– 16
8 –2
– 19
4
– 15
– 30
30
0
So, since the remainder is zero, – 2 is a zero of f(x). In fact, we can
rewrite f(x) as: 8x3 + 14x2 – 19x – 30 = (x + 2)(8x2 – 2x – 15)
Ex. 10
f(x) = x4 – 6x3 + 2x2 + 57x – 90;
Solution:
Using synthetic division, we get:
3
1
1
–6
3
–3
2
–9
–7
57
– 21
36
k=3
– 90
108
18
So, 3 is not a zero of f(x).
Ex. 11
f(x) = 4x3 – 21x2 + 72x – 65;
Solution:
Using synthetic division, we get:
2 – 3i
4
4
– 21
8 – 12i
– 13 – 12i
k = 2 – 3i
72
– 62 + 15i
10 + 15i
– 65
65
0
So, since the remainder is zero, 2 – 3i is a zero of f(x). We shall later
that if a complex number is a zero, its complex conjugate will also be
a zero.