1. No category

CLASS NOTES
SYNTHETIC DIVISION
When factoring or evaluating polynomials we often find that it is convenient to divide a
polynomial by a linear (first degree) binomial of the form x − k where k is a real number. In
certain problems we must use trial-and-error to find a particular value of k. The resulting
process of repeated divisions can be tedious and time-consuming. The use of synthetic division
(instead of long division) can save a lot of time and effort.
It must be emphasized that synthetic division may only be used when dividing a polynomial
by a binomial of the form x − k .
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Illustration: Suppose we wish to divide the polynomial P x = 5x 3 − 6 x 2 − 28 x − 2 by x − 3 .
The k-value is 3 (the number subtracted from x in the binomial x − 3 ). P x is a polynomial of
degree 3, and the four coefficients of P x are 5, − 6, − 28 and − 2 .
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The division may now be expressed as shown below. Note that there is a blank space above the
horizontal line.
3
g
5
−6
− 28
−2
We “bring down” (copy) the leading coefficient below the line in the same column.
3
g
5
B
−6
− 28
−2
5
We now multiply 5 (the number below the line in the first column) by the k-value (3) and write
the result in the next column (second column) above the line.
3
g
5
−6
− 28
−2
15
5
Then we add the two numbers that are above the line in the second column and write the sum
below the line.
3
g
5
−6
− 28
−2
15
5
9
We now multiply 9 (the number below the line in the second column) by the k-value (3) and
write the result in the next column (third column) above the line.
Page 1 of 10
3
g
5
−6
− 28
15
27
5
−2
9
Then we add the two numbers that are above the line in the third column and write the sum
below the line.
3
g
5
−6
− 28
15
27
9
−1
5
−2
We repeat the previous two steps one last time to get the final result.
3
g
5
−6
− 28
−2
15
27
−3
9
−1
−5
5
b
g
The first three numbers below the line 5, 9 and − 1 are the coefficients of the quotient. The
last number below the line −5 is the remainder R. We can therefore rewrite the division as the
sum of a polynomial (the quotient) and a rational (the remainder divided by the divisor).
b g
5x 3 − 6 x 2 − 28 x − 2
−5
= 5x 2 + 9 x − 1 +
x−3
x−3
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NOTE: We can see that P 3 = 5 3 − 6 3 − 28 3 − 2 = − 5 . I.e. the value of P k is equal
to the remainder, R, when we divide P x by x − k . More on this later (Remainder Theorem).
3
Example: Perform the division
2
3x 4 − 4 x 2 + 9 x − 6
and rewrite as a polynomial plus rational.
x+2
b g
NOTE: The divisor may be written x − −2
→ k = −2 . Furthermore, the numerator, which
is a polynomial of degree 4, has five terms. The five terms are 3x 4 + 0 x 3 − 4 x 2 + 9 x − 6 . We
use synthetic division to find the quotient and remainder.
−2
g3
0
−6
3
−6
−4
9
−6
12 − 16
14
−7
8
8
→
→ The quotient is 3x 3 − 6 x 2 + 8 x − 7 and the remainder is 8.
3x 4 − 4 x 2 + 9 x − 6
8
= 3x 3 − 6 x 2 + 8 x − 7 +
x+2
x+2
Page 2 of 10
Exercises: Perform each division and rewrite as a polynomial plus a rational.
x 3 + 7 x 2 + 15
1.
x+4
2x4 + 7x3 − x + 5
2.
x+3
4 x 4 − 7 x 3 + 9 x 2 − 3x − 8
3.
x −1
3x 5 − 4 x + 1
4.
x−2
Answers to Exercises:
−4
g1
7
0
15
− 4 − 12 48
1.
3 − 12 63
1
−3
g2
7
2.
g4
0
1 −3
8 − 19
4 −3
2
g3
0
6
0
6
→
3
3 −5
0 −4
3 6 12 24 44
89
−19
2x4 + 7x3 − x + 5
= 2 x 3 + x 2 − 3x + 8 +
x+3
x+3
−5
4 x 4 − 7 x 3 + 9 x 2 − 3x − 8
= 4 x 3 − 3x 2 + 6 x + 3 +
x −1
x −1
1
6 12 24 48 88
4.
→
9 − 3 −8
4 −3
x 3 + 7 x 2 + 15
63
= x 2 + 3x − 12 +
x+4
x+4
5
9 − 24
−7
3.
−1
−6 −3
2
1
→
→
3x 5 − 4 x + 1
89
= 3x 4 + 6 x 3 + 12 x 2 + 24 x + 44 +
x−2
x−2
REMAINDER THEOREM
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At the end of the previous illustration we saw that P 3 = −5 . Similarly, in the example we saw
that P −2 = 8 . In both cases the value of the polynomial at x = k is equal to the remainder R.
The same principle applies in each of the exercises above. This principle illustrates the
Remainder Theorem.
b g
Remainder Theorem:
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If the polynomial P x is divided by x − k , then the remainder R is equal to P k .
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Pk =R
Page 3 of 10
One application of the Remainder Theorem is the evaluation of polynomials at given k-values.
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Illustration: Suppose we wish to evaluate the polynomial P x = 2 x 4 + x 3 − 3x 2 + 5x + 4 at
k = −7 . Typically a student will find P −7 by replacing each x in the equation for P by −7 .
b g
Pb −7g = 2b −7g + b −7g − 3b −7g + 5b −7g + 4 or Pb −7g = 4281 . Even with the help of
4
3
2
This gives
a calculator this can be a tedious and frustrating calculation, especially if the number of terms is
large. If the student has learned Synthetic Division and is comfortable with it, then he/she will
find that the Remainder Theorem is a faster and more reliable way to find P −7 .
b g
g2
−7
−3
1
5
4
− 14
91 − 616
4277
2 − 13
88 − 611
4281
→
b g
P −7 = 4281
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Exercises: In each of the following evaluate the polynomial P x at the given k-value.
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3. Pb x g = 6 x
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4. Pb x g = 4 x + 2 x
1. P x = 2 x 3 − 8 x 2 − 4 x + 5 at k = 6
4
2. P x = 5x 4 − 3x 2 + 2 x − 1 at k = −4
− x 3 + 9 x 2 − 7 at k = 7
5
4
− 5x 3 + 6 x 2 − 3x − 8 at k = −5
Answers to Exercises:
6
g2
−8 −4
1.
2
−4
12
24 120
4
20 125
g5
0
−3
bg
→
2
P 6 = 125
−1
− 20 80 − 308 1224
2.
5 − 20
7
g6
−1
−5
9
6
41 296
g4
2
− 20
b g
→
P −4 = 1223
→
P 7 = 14497
77 − 306 1223
−7
0
42 287 2072 14504
3.
4.
5
−5
bg
2072 14497
6
−3
−8
90 − 425 2095 − 10460
→
4 − 18 85 − 419 2092 − 10468
Page 4 of 10
b g
P −5 = −10468
FACTOR THEOREM
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Pb x g
= Qb x g +
bg
We have seen that if a polynomial P x is divided by x − k to get the quotient Q x and
remainder R, then we can write
bg
x−k
R
. And we know, by the Remainder
x−k
Pb x g
= Qb x g . This last
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x−k
Pb x g = b x − k g ⋅ Qb x g which means that x − k is a factor of Pb x g .
Theorem, that P k = R . It follows that if P k = 0 then R = 0 and
equation may be rewritten
Factor Theorem:
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The polynomial x − k is a factor of polynomial P x if and only if P k = 0 .
b g
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Example: Given that P −3 = 0 rewrite the polynomial P x = 6 x 3 + 19 x 2 + 2 x − 3 as the
product of two factors.
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Solution: By the Factor Theorem we know that x + 3 is one factor of P x . To get the other
factor we can use synthetic division to divide P x by x + 3 (keeping in mind that k = −3 ).
−3
g6
6
bg
2 −3
19
− 18 − 3
3
1 −1
0
→
b g b gc
h
P x = x + 3 6x 2 + x − 1
bg
Exercises: In each of the following rewrite the polynomial P x as the product of two factors.
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Pb x g = 2 x + x
b g
1. P x = x 4 + 2 x 3 − 7 x 2 − 20 x − 12 and given P −2 = 0
2.
5
bg
− 30 x 3 − 35x 2 + 38 x + 24 and given P 4 = 0
4
Answers to Exercises:
−2
g1
2 − 7 − 20 − 12
−2
1.
1
4
g2
2
14
12
0 −7
−6
0
1 − 30 − 35
8
2.
0
9
36
38
→
24
24 − 44 − 24
6 − 11 − 6
b g b gc
h
P x = x + 2 x3 − 7x − 6
→
b g b gc
h
P x = x − 4 2 x 4 + 9 x 3 + 6 x 2 − 11x − 6
0
Page 5 of 10
FACTORING POLYNOMIALS OF HIGHER DEGREE (degree higher than 2)
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Terminology: A zero of a polynomial P is a number k such that P k = 0 (we worked with
“zeros” in the preceding section). A zero of P is called a root or solution of the polynomial
equation P x = 0 . Clearly the technical difference between “zeros” and “roots” is of little
importance to us. Furthermore, the term “zero” can be confusing to the student (a “zero” of the
polynomial is not necessarily equal to zero). Therefore, in the remainder of this discussion the
term “root” will be used exclusively.
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The Rational Roots Test:
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Suppose the polynomial P x has integer coefficients. If the rational number dn (in
lowest terms) is a root of P then the numerator n is a factor of the constant term of P
and the denominator d is a factor of the leading coefficient.
The Integer Roots Test:
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Suppose the polynomial P x has integer coefficients with leading coefficient 1. If the
integer k is a root of P then k is a factor of the constant term of polynomial P.
The Rational Roots Test and the Integer Roots Test can only give us possible roots of a given
polynomial. We must test these possible values (using synthetic division) to determine which, if
any, of the possibilities are actual roots. Once we have the actual root(s) then we can write the
polynomial in factored form.
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Example 1: Completely factor the polynomial P x = x 3 − 5x 2 − 2 x + 24 .
Solution: Polynomial P has integer coefficients with leading coefficient 1. We can therefore
apply the Integer Roots Test. Any possible integer root of P must be a factor of the constant
term 24. The possible integer roots are 1, 2, 3, etc. as well as their negatives. The complete list
can be written ±1, ± 2, ± 3, ± 4, ± 6, ± 12, ± 24 . This gives a total of 14 possible integer roots!
At most three of these can be actual roots. It is also possible that none of them are actual roots.
They must be tested one by one using synthetic division (trial-and-error).
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If we try k =1 we get a non-zero remainder which is, as we know, P 1 .
1
g1
− 5 − 2 − 24
1 −4
−6
→
bg
P 1 = −30
→
bg
P1 ≠0
1 − 4 − 6 − 30
Therefore k =1 is not a root of P. Similarly k = −1 and k = 2 are not roots of P.
However, as shown below, k = −2 is a root of P. It follows, by the Factor theorem, that x + 2 is
one of the factors of P.
Page 6 of 10
−2
g1
− 5 − 2 − 24
− 2 14
24
1 − 7 12
0
b g b gc
→
P x = x + 2 x 2 − 7 x + 12
h
We see that the other factor, in addition to x + 2 , is x 2 − 7 x + 12 . This second factor can itself be
factored using the techniques of High School algebra. Or it can be factored by continuing to test
the possible integer roots. The final, factored, form is P x = x + 2 x − 3 x − 4 .
b g b gb gb g
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Example 2: Completely factor the polynomial P x = x 5 − 26 x 3 + 12 x 2 + 97 x + 60 .
Solution: The possible integer roots are ±1, ± 2, ± 3, ± 4, ± 5, ± 6, ± 10, etc. (applying the
Integer Roots Test). We find that k = −1 is a root of P.
−1
g1
0 − 26 12
−1
1
97
60
1 − 1 − 25 37
60
b g b gc
→
25 − 37 − 60
h
P x = x + 1 x 4 − x 3 − 25x 2 + 37 x + 60
0
Working with the other (second) factor we test k = −1 a second time (roots can repeat!) and find
that it is also a root of the second factor.
−1
g1
− 1 − 25 37
−1
2
60
23 − 60
1 − 2 − 23 60
→
b g b gb gc
h
P x = x + 1 x + 1 x 3 − 2 x 2 − 23x + 60
0
Working with the third factor we find that k = 3 is a root.
3
g1
− 2 − 23
3
1
60
3 − 60
1 − 20
b g b g b x − 3gc x
→
P x = x +1
2
2
h
+ x − 20
0
As in Example 1, the quadratic factor x 2 + x − 20 can be factored by using the techniques of
High School algebra or by continuing to test the possible integer roots. The factored form is
b g b g b x − 3gb x + 5gb x − 4g .
P x = x +1
2
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Example 3: Completely factor the polynomial P x = 6 x 4 − 5x 3 + 26 x 2 − 25x − 20 .
Solution: Because the leading coefficient is not equal to 1, we apply the Rational Roots Test.
Suppose that dn is a rational root of P.
Then the possible values of the numerator n are ±1, ± 2, ± 4, ± 5, ± 10, ± 20 .
The possible values of the denominator d are ±1, ± 2, ± 3, ± 6 .
Page 7 of 10
We find the possible combinations (n divided by d ) of the above numbers. The possible rational
roots are ±1, ± 2, ± 4, ± 5, ± 10, ± 20, ± 21 , ± 25 , ± 13 , ± 23 , ± 43 , ± 53 , ± 10
, ± 20
, ± 16 , ± 56 .
3
3
We find that k = − 21 is a root of P.
−1
2
j6
− 5 26 − 25 − 20
−3
4 − 15
20
6 −8
30 − 40
0
b g e jc
→
h
P x = x + 21 6 x 3 − 8 x 2 + 30 x − 40
Working with the second factor we find that k = 43 is a root.
4
3
j6
−8
6
30 − 40
8
0
40
0
30
0
b g e je jc
→
h
P x = x + 21 x − 43 6 x 2 + 30
c
h
The quadratic factor 6 x 2 + 30 can be rewritten as 6 x 2 + 5 . This quadratic factor can not be
factored further. The final (simplified) factored form is
b g e je jc
h
P x = 6 x + 21 x − 43 x 2 + 5
bg b
gb
gc
h
which can be rewritten as P x = 2 x + 1 3x − 4 x 2 + 5 .
Exercises: Completely factor each of the following polynomials.
bg
Pb x g = 4 x
1. P x = x 4 − 27 x 2 − 14 x + 120
3.
4
bg
Pb x g = 3x + 10 x
2. P x = x 4 + x 3 − 15x 2 + 23x − 10
− 9 x 3 − 25x 2 + 36 x + 36
4.
5
4
− 11x 3 − 46 x 2 − 4 x + 24
Solutions:
1. The possible integer roots are ±1, ± 2, ± 3, ± 4, ± 5, ± 6, ± 10, etc. (Integer Roots Test). We
test these possibilities (using synthetic division) to find the actual roots of the polynomial.
2
g1
0 − 27 − 14
2
120
4 − 46 − 120
1 2 − 23 − 60
b g b gc
h
→ k = 2 is a root → P x = x − 2 x 3 + 2 x 2 − 23x − 60
0
We continue by testing for roots in the second factor.
−3
g1
2 − 23 − 60
−3
3
60
1 − 1 − 20
0
b g b gb gc
h
→ k = −3 is a root → P x = x − 2 x + 3 x 2 − x − 20
The third factor may be factored by the techniques of High school algebra.
b g b gb gb gb g
P x = x −2 x + 3 x + 4 x −5
Page 8 of 10
2. The possible integer roots are ±1, ± 2, ± 5, ± 10 (Integer Roots Test). We test these
possibilities to find the actual roots of the polynomial.
1
g1
1 − 15
23 − 10
2 − 13
1
1 2 − 13
b g b gc
10
h
→ k = 1 is a root → P x = x − 1 x 3 + 2 x 2 − 13x + 10
10
0
We continue testing for roots in the second factor. As mentioned earlier, it is important to
test the same root a second time. Roots can, and sometimes do, repeat.
1
g1
2 − 13
10
b g b gb gc
3 − 10
1
1 3 − 10
h
→ k = 1 is a root (again) → P x = x − 1 x − 1 x 2 + 3x − 10
0
The third factor may be factored by the techniques of High school algebra.
b g b g b x + 5gb x − 2g
P x = x −1
2
3. The Rational Roots Test allows us to find the possible rational roots:
±1, ± 2, ± 3, ± 4, ± 6, ± 9, ± 12, ± 18, ± 36, ± 21 , ± 23 , ± 92 , ± 41 , ± 43 , ± 94
We test these possibilities to find the actual roots of the polynomial.
g4
2
− 9 − 25
36
36
b g b gc
− 2 − 54 − 36
8
4 − 1 − 27 − 18
h
→ k = 2 is a root → P x = x − 2 4 x 3 − x 2 − 27 x − 18
0
We continue by testing for roots in the second factor.
−2
g4
− 1 − 27 − 18
−8
18
18
4 −9
−9
0
b g b gb gc
→ k = −3 is a root → P x = x − 2 x + 2 4 x 2 − 9 x − 9
h
The third factor may be factored by the techniques of High school algebra.
b g b gb gb gb
g
P x = x − 2 x + 2 x − 3 4x + 3
4. The possible rational roots are ±1, ± 2, ± 3, ± 4, ± 6, ± 8, ± 12, ± 24, ± 13 , ± 23 , ± 43 , ± 83 .
We test these possibilities to find the actual roots of the polynomial.
g
− 1 3 10 − 11 − 46 − 4
−3 −7
18
3 7 − 18 − 28
24
28 − 24
24
b g b gc
→ − 1 is a root → P x = x + 1 3x 4 + 7 x 3 − 18 x 2 − 28 x + 24
0
Page 9 of 10
h
We continue by testing for roots in the second factor.
2
g3
7 − 18 − 28
6
24
8 − 12
3 13
b g b gb gc
16 − 24
26
→ k = 2 is a root → P x = x + 1 x − 2 3x 3 + 13x 2 + 8 x − 12
h
0
We continue by testing for roots in the third factor.
−2
g3
8 − 12
13
− 6 − 14
12
−6
0
3
7
b g b gb gb gc
h
→ k = −2 is a root → P x = x + 1 x − 2 x + 2 3x 2 + 7 x − 6
The fourth factor may be factored by the techniques of High school algebra.
b g b gb gb gb
gb g
P x = x + 1 x − 2 x + 2 3x − 2 x + 3
Drill Exercises: Completely factor each of the following polynomials.
bg
3. Pb x g = x + 9 x − 6 x − 116 x + 168
5. Pb x g = x + 8 x + 5x − 74 x − 120
7. Pb x g = 2 x + x − 85x + 42
9. Pb x g = 20 x + 12 x − 33x − 5x + 6
11. Pb x g = 8 x − 20 x − 18 x + 81x − 54
1. P x = x 3 + 5x 2 − 12 x − 36
4
3
2
4
3
2
3
2
4
4
3
2
3
2
Answers to Drill Exercises:
b g b gb gb g
3. Pb x g = b x + 7gb x + 6gb x − 2g
5. Pb x g = b x + 5gb x + 4gb x + 2gb x − 3g
7. Pb x g = b x + 7gb2 x − 1gb x − 6g
9. Pb x g = b2 x + 3gb2 x + 1gb5x − 2gb x − 1g
11. Pb x g = b x + 2gb2 x − 3g
1. P x = x + 6 x + 2 x − 3
2
3
bg
4. Pb x g = x − 6 x − 15x + 80 x + 84 x − 144
6. Pb x g = x + 6 x − 28 x − 110 x − 117 x − 40
8. Pb x g = 24 x − 62 x + 31x + 12
10. Pb x g = 9 x + 30 x − 92 x − 498 x − 685x − 300
12. Pb x g = 72 x + 90 x − 383x − 287 x + 254 x − 40
2. P x = x 3 − 6 x 2 − 7 x + 60
5
4
3
5
4
3
3
5
2
2
4
5
2
3
2
4
3
2
b g b gb gb g
4. Pb x g = b x + 3gb x + 2gb x − 1gb x − 4gb x − 6g
6. Pb x g = b x + 8gb x − 5gb x + 1g
8. Pb x g = b4 x + 1gb2 x − 3gb3x − 4g
10. Pb x g = b x + 3gb x + 1gb x − 4gb3x + 5g
12. Pb x g = b2 x + 5gb3x + 4gb4 x − 1gb3x − 1gb x − 2g
2. P x = x + 3 x − 4 x − 5
3
2
Page 10 of 10