OpenStax-CNX module: m21904
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Factoring Polynomials: Factoring
Trinomials with Leading
∗
Coefficient 1
Wade Ellis
Denny Burzynski
This work is produced by OpenStax-CNX and licensed under the
Creative Commons Attribution License 3.0†
Abstract
This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. Factoring is an
essential skill for success in algebra and higher level mathematics courses. Therefore, we have taken
great care in developing the student's understanding of the factorization process. The technique is
consistently illustrated by displaying an empty set of parentheses and describing the thought process
used to discover the terms that are to be placed inside the parentheses. The factoring scheme for special
products is presented with both verbal and symbolic descriptions, since not all students can interpret
symbolic descriptions alone. Two techniques, the standard "trial and error" method, and the "collect
and discard" method (a method similar to the "ac" method), are presented for factoring trinomials with
leading coecients dierent from 1. Objectives of this module: be able to factor trinomials with leading
coecient 1, become familiar with some factoring hints.
1 Overview
• Method
• Factoring Hints
2 Method
Let's consider the product of the two binomials (x + 4) and (x + 7).
Notice that the rst term in the resulting trinomial comes from the product of the rst terms in
the binomials: x · x = x2 . The last term in the trinomial comes from the product of the last terms in
the binomials: 4 · 7 = 28. The middle term comes from the addition of the outer and inner products:
∗
†
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OpenStax-CNX module: m21904
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7x + 4x = 11x. Also, notice that the coecient of the middle term is exactly the
the binomials: 4 + 7 = 11.
sum of the last terms in
The problem we're interested in is that given a trinomial, how can we nd the factors? When the leading
coecient (the coecient of the quadratic term) is 1, the observations we made above lead us to the following
method of factoring.
Method of Factoring
1. Write two sets of parentheses:
.
2. Place a binomial into each set of parentheses. The rst term of each binomial is a factor of the rst
term of the trinomial.
3. Determine the second terms of the binomials by determining the factors of the third term that when
added together yield the coecient of the middle term.
3 Sample Set A
Factor the following trinomials.
Example 1
x2 + 5x + 6
1. Write two sets of parentheses:
.
2
2. Place the factors of x into the rst position of each set of parentheses:
(x
(x
3. The third term of the trinomial is 6. We seek two numbers whose
(a) product is 6 and
(b) sum is 5.
The required numbers are 3 and 2. Place +3 and + 2 into the parentheses.
x2 + 5x + 6 = (x + 3) (x + 2)
The factorization is complete. We'll check to be sure.
(x + 3) (x + 2)
=
x2 + 2x + 3x + 6
= x2 + 5x + 6
Example 2
y 2 − 2y − 24
1. Write two sets of parentheses:
.
2
2. Place the factors of y into the rst position of each set of parentheses:
(y
(y
3. The third term of the trinomial is −24. We seek two numbers whose
(a) product is −24 and
(b) sum is −2.
The required numbers are −6 and 4. Place −6 and + 4 into the parentheses.
y 2 − 2y − 24 = (y − 6) (y + 4)
The factorization is complete. We'll check to be sure.
(y − 6) (y + 4)
=
y 2 + 4y − 6y − 24
= y 2 − 2y − 24
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Notice that the other combinations of the factors of −24(some of which are −2, 12; 3, −8; and − 4, 6)
do not work. For example,
(y − 2) (y + 12) = y 2 + 10 y − 24
(y + 3) (y − 8)
=
y 2 − 5 y − 24
(y − 4) (y + 6)
=
y 2 + 2 y − 24
In all of these equations, the middle terms are incorrect.
Example 3
a2 − 11a + 30
1. Write two sets of parentheses:
.
2
2. Place the factors of a into the rst position of each set of parentheses:
(a
(a
3. The third term of the trinomial is +30. We seek two numbers whose
(a) product is 30 and
(b) sum is −11.
The required numbers are −5 and − 6. Place −5 and − 6into the parentheses.
a2 − 11a + 30 = (a − 5) (a − 6)
The factorization is complete. We'll check to be sure.
(a − 5) (a − 6)
=
a2 − 6a − 5a + 30
=
a2 − 11a + 30
Example 4
3x2 − 15x − 42
Before we begin, let's recall the most basic rule of factoring: factor out common monomial
factors rst. Notice that 3 is the greatest
common monomial factor of every term. Factor out 3.
3x2 − 15x − 42 = 3 x2 − 5x − 14
Now we can continue.
1. Write two sets of parentheses:3
.
2
2. Place the factors of x into the rst position of each set of parentheses:
3 (x
(x
3. The third term of the trinomial is −14. We seek two numbers whose
(a) product is −14 and
(b) sum is −5.
The required numbers are −7 and 2. Place −7 and + 2into the parentheses.
3x2 − 15x − 42 = 3 (x − 7) (x + 2)
The factorization is complete. We'll check to be sure.
3 (x − 7) (x + 2)
3 x2 + 2x − 7x − 14
= 3 x2 − 5x − 14
=
=
3x2 − 15x − 42
4 Practice Set A
Factor, if possible, the following trinomials.
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Exercise 1
(Solution on p. 8.)
Exercise 2
(Solution on p. 8.)
Exercise 3
(Solution on p. 8.)
Exercise 4
(Solution on p. 8.)
k 2 + 8k + 15
y 2 + 7y − 30
m2 + 10m + 24
m2 − 10m + 16
5 Factoring Hints
Factoring trinomials may take some practice, but with time and experience, you will be able to factor much
more quickly.
There are some clues that are helpful in determining the factors of the third term that when added yield
the coecient of the middle term.
Factoring Hints
Look at the sign of the last term:
a. If the sign is positive, we know that the two factors must have the same sign, since (+) (+) =
(+) and (−) (−) = (+). The two factors will have the same sign as the sign of the middle term.
b. If the sign is negative, we know that two factors must have opposite signs, since (+) (−) = (−) and (−) (+) =
(−).
6 Sample Set B
Example 5
Factor x2 − 7x + 12.
1. Write two sets of parentheses:
.
2. The third term of the trinomial is +12. The sign is positive, so the two factors of 12 we are
looking for must have the same sign. They will have the sign of the middle term. The sign
of the middle term is negative, so both factors of 12 are negative. They are −12 and − 1, −
6 and − 2, or − 4 and − 3. Only the factors −4 and − 3 add to − 7, so − 4 and − 3 are the
proper factors of 12 to be used.
x2 − 7x + 12 = (x − 4) (x − 3)
7 Practice Set B
Factor, if possible, the following trinomials.
Exercise 5
(Solution on p. 8.)
Exercise 6
(Solution on p. 8.)
Exercise 7
(Solution on p. 8.)
4k 2 + 32k + 28
3y 4 + 24y 3 + 36y 2
x2 − xy − 6y 2
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OpenStax-CNX module: m21904
Exercise 8
5
(Solution on p. 8.)
−5a5 b − 10a4 b2 + 15a3 b3
8 Exercises
For the following problems, factor the trinomials when possible.
Exercise 9
(Solution on p. 8.)
x2 + 4x + 3
Exercise 10
x2 + 6x + 8
Exercise 11
(Solution on p. 8.)
x2 + 7x + 12
Exercise 12
x2 + 6x + 5
Exercise 13
(Solution on p. 8.)
y 2 + 8y + 12
Exercise 14
y 2 − 5y + 6
Exercise 15
(Solution on p. 8.)
y 2 − 5y + 4
Exercise 16
a2 + a − 6
Exercise 17
(Solution on p. 8.)
a2 + 3a − 4
Exercise 18
x2 + 4x − 21
Exercise 19
(Solution on p. 8.)
x2 − 4x − 21
Exercise 20
x2 + 7x + 12
Exercise 21
(Solution on p. 8.)
y 2 + 10y + 16
Exercise 22
x2 + 6x − 16
Exercise 23
(Solution on p. 8.)
y 2 − 8y + 7
Exercise 24
y 2 − 5y − 24
Exercise 25
(Solution on p. 8.)
a2 + a − 30
Exercise 26
a2 − 3a + 2
Exercise 27
a2 − 12a + 20
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(Solution on p. 8.)
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Exercise 28
y 2 − 4y − 32
Exercise 29
(Solution on p. 8.)
x2 + 13x + 42
Exercise 30
x2 + 2x − 35
Exercise 31
(Solution on p. 8.)
x2 + 13x + 40
Exercise 32
y 2 + 6y − 27
Exercise 33
(Solution on p. 8.)
b2 + 15b + 56
Exercise 34
3a2 + 24a + 36
(Hint: Always search for a common factor.)
Exercise 35
(Solution on p. 8.)
4x2 + 12x + 8
Exercise 36
2a2 − 18a + 40
Exercise 37
(Solution on p. 8.)
5y 2 − 70y + 440
Exercise 38
6x2 − 54x + 48
Exercise 39
(Solution on p. 8.)
x3 + 6x2 + 8x
Exercise 40
x3 − 8x2 + 15x
Exercise 41
(Solution on p. 8.)
x4 + 9x3 + 14x2
Exercise 42
2a3 + 12a2 + 10a
Exercise 43
(Solution on p. 9.)
4a3 − 40a2 + 84a
Exercise 44
3xm2 + 33xm + 54x
Exercise 45
(Solution on p. 9.)
2y 2 n2 − 10y 2 n − 48y 2
Exercise 46
4x4 − 42x3 + 144x2
Exercise 47
y 5 + 13y 4 + 42y 3
Exercise 48
4x2 a6 − 48x2 a5 + 252x2 a4
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(Solution on p. 9.)
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9 Exercises for Review
Exercise 49
() Factor 6xy + 2ax − 3ay − a2 .
Exercise 50
() Factor 8a2 − 50.
Exercise 51
() Factor 4x2 + 17x − 15.
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(Solution on p. 9.)
(Solution on p. 9.)
OpenStax-CNX module: m21904
Solutions to Exercises in this Module
Solution to Exercise (p. 3)
(k + 3) (k + 5)
Solution to Exercise (p. 4)
(y + 10) (y − 3)
Solution to Exercise (p. 4)
(m + 6) (m + 4)
Solution to Exercise (p. 4)
(m − 8) (m − 2)
Solution to Exercise (p. 4)
4 (k + 7) (k + 1)
Solution to Exercise (p. 4)
3y 2 (y + 2) (y + 6)
Solution to Exercise (p. 4)
(x + 2y) (x − 3y)
Solution to Exercise (p. 5)
−5a3 b (a + 3b) (a − b)
Solution to Exercise (p. 5)
(x + 3) (x + 1)
Solution to Exercise (p. 5)
(x + 3) (x + 4)
Solution to Exercise (p. 5)
(y + 6) (y + 2)
Solution to Exercise (p. 5)
(y − 4) (y − 1)
Solution to Exercise (p. 5)
(a + 4) (a − 1)
Solution to Exercise (p. 5)
(x − 7) (x + 3)
Solution to Exercise (p. 5)
(y + 8) (y + 2)
Solution to Exercise (p. 5)
(y − 7) (y − 1)
Solution to Exercise (p. 5)
(a + 6) (a − 5)
Solution to Exercise (p. 5)
(a − 10) (a − 2)
Solution to Exercise (p. 6)
(x + 6) (x + 7)
Solution to Exercise (p. 6)
(x + 5) (x + 8)
Solution to Exercise (p. 6)
(b + 8) (b + 7)
Solution to Exercise (p. 6)
4 (x + 2) (x + 1)
Solution to Exercise
(p. 6)
5 y 2 − 14y + 88
Solution to Exercise (p. 6)
x (x + 4) (x + 2)
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OpenStax-CNX module: m21904
Solution to Exercise (p. 6)
x2 (x + 7) (x + 2)
Solution to Exercise (p. 6)
4a (a − 7) (a − 3)
Solution to Exercise (p. 6)
2y 2 (n − 8) (n + 3)
Solution to Exercise (p. 6)
y 3 (y + 6) (y + 7)
Solution to Exercise (p. 7)
(2x − a) (3y + a)
Solution to Exercise (p. 7)
(4x − 3) (x + 5)
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