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Section7.3Factoring Trinomials with Leading Coefficient One

ObjectivesPCC Course Content and Outcome Guide

In Chapter 6, we learned how to multiply binomials like (x+2)(x+3) and obtain the trinomial x2+5x+6. In this section, we will learn how to undo that. So we'll be starting with a trinomial like x2+5x+6 and obtaining its factored form (x+2)(x+3). The trinomials that we'll factor in this section all have leading coefficient 1, but Section 7.4 will cover some more general trinomials.

Figure7.3.1Alternative Video Lesson

Subsection7.3.1Factoring Trinomials by Listing Factor Pairs

Consider the example x2+5x+6=(x+2)(x+3). There are at least three things that are important to notice:

  • The leading coefficient of x2+5x+6 is 1.

  • The two factors on the right use the numbers 2 and 3, and when you multiply these you get the 6 from x2+5x+6.

  • The two factors on the right use the numbers 2 and 3, and when you add these you get the 5 from x2+5x+6.

So the idea is that if you need to factor x2+5x+6 and you somehow discover that 2 and 3 are special numbers (because 2β‹…3=6 and 2+3=5), then you can conclude that (x+2)(x+3) is the factored form of the given polynomial.

Example7.3.2

Factor x2+13x+40. Since the leading coefficient is 1, we are looking to write this polynomial as (x+?)(x+?) where the question marks are two possibly different, possibly negative, numbers. We need these two numbers to multiply to 40 and add to 13. How can you track these two numbers down? Since the numbers need to multiply to 40, one method is to list all factor pairs of 40 in a table just to see what your options are. We'll write every pair of factors that multiply to 40.

1β‹…40
2β‹…20
4β‹…10
5β‹…8
βˆ’1β‹…(βˆ’40)
βˆ’2β‹…(βˆ’20)
βˆ’4β‹…(βˆ’10)
βˆ’5β‹…(βˆ’8)

We wanted to find all factor pairs. To avoid missing any, we started using 1 as a factor, and then slowly increased that first factor. The table skips over using 3 as a factor, because 3 is not a factor of 40. Similarly the table skips using 6 and 7 as a factor. And there would be no need to continue with 8 and beyond, because we already found β€œlarge” factors like 8 as the partners of β€œsmall” factors like 5.

There is an entire second column where the signs are reversed, since these are also ways to multiply two numbers to get 40. In the end, there are eight factor pairs.

We need a pair of numbers that also adds to 13. All we need to do is check what each of our factor pairs add up to:

Factor Pair Sum of the Pair
1β‹…40 41
2β‹…20 22
4β‹…10 14
5β‹…8 13 (what we wanted)
Factor Pair Sum of the Pair
βˆ’1β‹…(βˆ’40) (no need to go this far)
βˆ’2β‹…(βˆ’20) (no need to go this far)
βˆ’4β‹…(βˆ’10) (no need to go this far)
βˆ’5β‹…(βˆ’8) (no need to go this far)

The winning pair of numbers is 5 and 8. Again, what matters is that 5β‹…8=40, and 5+8=13. So we can conclude that x2+13x+40=(x+5)(x+8).

To ensure that we made no mistakes, here are two possible checks.

Multiply it Out

Multiplying out our answer (x+5)(x+8) should give us x2+13x+40:

(x+5)(x+8)=(x+5)β‹…x+(x+5)β‹…8=x2+5x+8x+40=βœ“x2+13x+40

We can also use a rectangular area diagram to verify the factorization is correct:

x 5
x x2 5x
8 8x 40
Evaluating

If the answer really is (x+5)(x+8), then notice how evaluating at βˆ’5 would result in 0. So the original expression should also result in 0 if we evaluate at βˆ’5. And similarly, if we evaluate it at βˆ’8, x2+13x+40 should be 0.

(βˆ’5)2+13(βˆ’5)+40=?0(βˆ’8)2+13(βˆ’8)+40=?025βˆ’65+40=?064βˆ’104+40=?00=βœ“00=βœ“0.

This also gives us evidence that the factoring was correct.

Example7.3.3

Factor y2βˆ’11y+24. The negative coefficient is a small complication from Example 7.3.2, but the process is actually still the same.

Solution

We need a pair of numbers that multiply to \(24\) and add to \(-11\text{.}\) Note that we do care to keep track that they sum to a negative total.

Factor Pair Sum of the Pair
\(1\cdot24\) \(25\)
\(2\cdot12\) \(14\)
\(3\cdot8\) \(11\) (close; wrong sign)
\(4\cdot6\) \(10\)
Factor Pair Sum of the Pair
\(-1\cdot(-24)\) \(-25\)
\(-2\cdot(-12)\) \(-14\)
\(-3\cdot(-8)\) \(-11\) (what we wanted)
\(-4\cdot(-6)\) (no need to go this far)

So \(y^2-11y+24=(y-3)(y-8)\text{.}\) To confirm that this is correct, we should check. Either by multiplying out the factored form:

\begin{align*} (y-3)(y-8)\amp=(y-3)\cdot y-(y-3)\cdot8\\ \amp=y^2-3y-8y+24\\ \amp\stackrel{\checkmark}{=}y^2-11y+24 \end{align*}
\(y\) \(-3\)
\(y\) \(y^2\) \(-3y\)
\(-8\) \(-8y\) \(24\)

Or by evaluating the original expression at \(3\) and \(8\text{:}\)

\begin{align*} 3^2-11(3)+24\amp\stackrel{?}{=}0\amp8^2-11(8)+24\amp\stackrel{?}{=}0\\ 9-33+24\amp\stackrel{?}{=}0\amp 64-88+24\amp\stackrel{?}{=}0\\ 0\amp\stackrel{\checkmark}{=}0\amp 0\amp\stackrel{\checkmark}{=}0\text{.} \end{align*}

Our factorization passes the tests.

Example7.3.4

Factor z2+5zβˆ’6. The negative coefficient is again a small complication from Example 7.3.2, but the process is actually still the same.

Solution

We need a pair of numbers that multiply to \(-6\) and add to \(5\text{.}\) Note that we do care to keep track that they multiply to a negative product.

Factor Pair Sum of the Pair
\(1\cdot(-6)\) \(-5\) (close; wrong sign)
\(2\cdot(-3)\) \(14\)
Factor Pair Sum of the Pair
\(-1\cdot6\) \(5\) (what we wanted)
\(-2\cdot3\) (no need to go this far)

So \(z^2+5z-6=(z-1)(z+6)\text{.}\) To confirm that this is correct, we should check. Either by multiplying out the factored form:

\begin{align*} (z-1)(z+6)\amp=(z-1)\cdot z+(z-1)\cdot6\\ \amp=z^2-z+6z-6\\ \amp\stackrel{\checkmark}{=}z^2+5z-6 \end{align*}
\(z\) \(-1\)
\(z\) \(z^2\) \(-z\)
\(6\) \(6z\) \(-6\)

Or by evaluating the original expression at \(1\) and \(-6\text{:}\)

\begin{align*} 1^2+5(1)-6\amp\stackrel{?}{=}0\amp(-6)^2+5(-6)-6\amp\stackrel{?}{=}0\\ 1+5-6\amp\stackrel{?}{=}0\amp 36-30-6\amp\stackrel{?}{=}0\\ 0\amp\stackrel{\checkmark}{=}0\amp 0\amp\stackrel{\checkmark}{=}0\text{.} \end{align*}

Our factorization passes the tests.

Checkpoint7.3.5

Try one as an exercise.

Subsection7.3.2Connection to Grouping

The factoring method we just learned takes a bit of a shortcut. To prepare yourself for a more complicated factoring technique in Section 7.4, you may want to try taking the β€œscenic route” instead of that shortcut.

Example7.3.6

Let's factor x2+13x+40 again (the polynomial from Example 7.3.2). As before, it is important to discover that 5 and 8 are important numbers, because they multiply to 40 and add to 13. As before, listing out all of the factor pairs is one way to discover the 5 and the 8.

Instead of jumping to the factored answer, we can show how x2+13x+40 factors in a more step-by-step fashion using 5 and 8. Since they add up to 13, we can write:

x2+13↓x+40=x2+5x+8xβžβ†“+40

We have intentionally split up the trinomial into an unsimplified polynomial with four terms. In Section 7.2, we handled such four-term polynomials by grouping:

=(x2+5x)+(8x+40)

Now we can factor out each group's greatest common factor:

=x(x+5)+8(x+5)=x(x+5)βžβ†“+8(x+5)βžβ†“=(x+5)(x+8)

And we have found that x2+13x+40 factors as (x+5)(x+8) without memorizing the shortcut.

This approach takes more time, and ultimately you may not use it much. However, if you try a few examples this way, it may make you more comfortable with the more complicated technique in Section 7.4.

Subsection7.3.3Trinomials with Higher Powers

So far we have only factored examples of quadratic trinomials: trinomials whose highest power of the variable is 2. However, this technique can also be used to factor trinomials where there is a larger highest power of the variable. It only requires that the highest power is even, that the next highest power is half of the highest power, and that the third term is a constant term.

In the four examples below, check:

  1. if the highest power is even

  2. if the next highest power is half of the highest power

  3. if the last term is constant

Factor pairs will help with…

  • y6βˆ’23y3βˆ’50

  • h16+22h8+105

Factor pairs won't help with…

  • y5βˆ’23y3βˆ’50

  • h16+22h8+105h2

Example7.3.7

Factor h16+22h8+105. This polynomial is one of the examples above where using factor pairs will help. We find that 7β‹…15=105, and 7+15=22, so the numbers 7 and 15 can be used:

h16+22h8+105=h16+7h8+15h8⏞+105=(h16+7h8)+(15h8+105)=h8(h8+7)+15(h8+7)=(h8+7)(h8+15)

Actually, once we settled on using 7 and 15, we could have concluded that h16+22h8+105 factors as (h8+7)(h8+15), if we know which power of h to use. We'll always use half the highest power in these factorizations.

In any case, to confirm that this is correct, we should check by multiplying out the factored form:

(h8+7)(h8+15)=(h8+7)β‹…h8+(h8+7)β‹…15=h16+7h8+15h8+105=βœ“h16+22h8+15
h8 7
h8 h16 7h8
15 15h8 105

Our factorization passes the tests.

Checkpoint7.3.8

Try one as an exercise.

Subsection7.3.4Factoring in Stages

Sometimes factoring a polynomial will take two or more β€œstages”. Always begin factoring a polynomial by factoring out its greatest common factor, and then apply a second stage where you use a technique from this section. The process of factoring a polynomial is not complete until each of the factors cannot be factored further.

Example7.3.9

Factor 2z2βˆ’6zβˆ’80.

Solution

We will first factor out the common factor, \(2\text{:}\)

\begin{equation*} 2z^2-6z-80=2\left(z^2-3z-40\right) \end{equation*}

Now we are left with a factored expression that might factor more. Looking inside the parentheses, we ask ourselves, β€œwhat two numbers multiply to be \(-40\) and add to be \(-3\text{?}\)” Since \(5\) and \(-8\) do the job the full factorization is:

\begin{align*} 2z^2-6z-80\amp=2\left(z^2-3z-40\right)\\ \amp=2(z+5)(z-8) \end{align*}
Example7.3.10

Factor βˆ’r2+2r+24.

Solution

The three terms don't exactly have a common factor, but as discussed in Section 7.1, when the leading term has a negative sign, it is often helpful to factor out that negative sign:

\begin{equation*} -r^2+2r+24=-\mathopen{}\left(r^2-2r-24\right)\mathclose{}\text{.} \end{equation*}

Looking inside the parentheses, we ask ourselves, β€œwhat two numbers multiply to be \(-24\) and add to be \(-2\text{?}\)” Since \(-6\) and \(4\) work here and the full factorization is shown:

\begin{align*} -r^2+2r+24\amp=-\mathopen{}\left(r^2-2r-24\right)\mathclose{}\\ \amp=-(r-6)(r+4) \end{align*}
Example7.3.11

Factor p2q3+4p2q2βˆ’60p2q.

Solution

First, always look for the greatest common factor: in this trinomial it is \(p^2q\text{.}\) After factoring this out, we have

\begin{equation*} p^2q^3+4p^2q^2-60p^2q=p^2q\mathopen{}\left(q^2+4q-60\right)\mathclose{}\text{.} \end{equation*}

Looking inside the parentheses, we ask ourselves, β€œwhat two numbers multiply to be \(-60\) and add to be \(4\text{?}\)” Since \(10\) and \(-6\) fit the bill, the full factorization can be shown below:

\begin{align*} p^2q^3+4p^2q^2-60p^2q\amp=p^2q\mathopen{}\left(q^2+4q-60\right)\mathclose{}\\ \amp=p^2q(q+10)(q-6) \end{align*}

Subsection7.3.5More Trinomials with Two Variables

You might encounter a trinomial with two variables that can be factored using the methods we've discussed in this section. It can be tricky though: x2+5xy+6y2 has two variables and it can factor using the methods from this section, but x2+5x+6y2 also has two variables and it cannot be factored. So in examples of this nature, it is even more important to check that factorizations you find actually work.

Example7.3.12

Factor x2+5xy+6y2. This is a trinomial, and the coefficient of x is 1, so maybe we can factor it. We want to write (x+?)(x+?) where the question marks will be something that makes it all multiply out to x2+5xy+6y2.

Since the last term in the polynomial has a factor of y2, it is natural to wonder if there is a factor of y in each of the two question marks. If there were, these two factors of y would multiply to y2. So it is natural to wonder if we are looking for (x+?y)(x+?y) where now the question marks are just numbers.

At this point we can think like we have throughout this section. Are there some numbers that multiply to 6 and add to 5? Yes, specifically 2 and 3. So we suspect that (x+2y)(x+3y) might be the factorization.

To confirm that this is correct, we should check by multiplying out the factored form:

(x+2y)(x+3y)=(x+2y)β‹…x+(x+2y)β‹…3y=x2+2xy+3xy+6y2=βœ“x2+5xy+6y2
x 2y
x x2 2xy
3y 3xy 6y2

Our factorization passes the tests.

In Section 7.4, there is a more definitive method for factoring polynomials of this form.

SubsectionExercises

1

Factor the given polynomial

r2+13r+30=

2

Factor the given polynomial

r2+14r+40=

3

Factor the given polynomial

t2+14t+48=

4

Factor the given polynomial

t2+5t+6=

5

Factor the given polynomial

x2+4xβˆ’45=

6

Factor the given polynomial

x2βˆ’4xβˆ’12=

7

Factor the given polynomial

y2+7yβˆ’18=

8

Factor the given polynomial

y2+2yβˆ’63=

9

Factor the given polynomial

y2βˆ’8y+15=

10

Factor the given polynomial

r2βˆ’9r+14=

11

Factor the given polynomial

r2βˆ’10r+16=

12

Factor the given polynomial

t2βˆ’11t+30=

13

Factor the given polynomial

t2+11t+10=

14

Factor the given polynomial

x2+12x+32=

15

Factor the given polynomial

x2+12x+32=

16

Factor the given polynomial

y2+3y+2=

17

Factor the given polynomial

y2+5yβˆ’14=

18

Factor the given polynomial

y2+5yβˆ’36=

19

Factor the given polynomial

r2βˆ’4rβˆ’60=

20

Factor the given polynomial

r2+4rβˆ’21=

21

Factor the given polynomial

t2βˆ’7t+12=

22

Factor the given polynomial

t2βˆ’17t+70=

23

Factor the given polynomial

x2βˆ’9x+14=

24

Factor the given polynomial

x2βˆ’9x+18=

25

Factor the given polynomial

y2βˆ’4y+7=

26

Factor the given polynomial

y2+3y+8=

27

Factor the given polynomial

y2βˆ’3y+8=

28

Factor the given polynomial

r2βˆ’6r+10=

29

Factor the given polynomial

r2+14r+49=

30

Factor the given polynomial

t2+6t+9=

31

Factor the given polynomial

t2+20t+100=

32

Factor the given polynomial

x2+12x+36=

33

Factor the given polynomial

x2βˆ’4x+4=

34

Factor the given polynomial

x2βˆ’20x+100=

35

Factor the given polynomial

y2βˆ’12y+36=

36

Factor the given polynomial

y2βˆ’4y+4=

37

Factor the given polynomial

4r2+4rβˆ’24=

38

Factor the given polynomial

2r2+4rβˆ’6=

39

Factor the given polynomial

8t2βˆ’8=

40

Factor the given polynomial

7t2βˆ’7=

41

Factor the given polynomial

3x2βˆ’15x+12=

42

Factor the given polynomial

3x2βˆ’15x+12=

43

Factor the given polynomial

2x2βˆ’20x+18=

44

Factor the given polynomial

6y2βˆ’24y+18=

45

Factor the given polynomial

7y5+28y4+21y3=

46

Factor the given polynomial

2r4+14r3+20r2=

47

Factor the given polynomial

3r4+15r3+12r2=

48

Factor the given polynomial

3t6+24t5+21t4=

49

Factor the given polynomial

3t6+18t5βˆ’21t4=

50

Factor the given polynomial

7x4+7x3βˆ’14x2=

51

Factor the given polynomial

9x10+9x9βˆ’18x8=

52

Factor the given polynomial

8x4βˆ’16x3βˆ’24x2=

53

Factor the given polynomial

10y7βˆ’30y6+20y5=

54

Factor the given polynomial

4y8βˆ’24y7+20y6=

55

Factor the given polynomial

2r9βˆ’14r8+12r7=

56

Factor the given polynomial

5r4βˆ’25r3+20r2=

57

Factor the given polynomial

βˆ’t2+7t+8=

58

Factor the given polynomial

βˆ’t2+3t+40=

59

Factor the given polynomial

βˆ’x2βˆ’4x+5=

60

Factor the given polynomial

βˆ’x2βˆ’5x+24=

61

Factor the given polynomial

x2+9xy+20y2=

62

Factor the given polynomial

y2+11yx+28x2=

63

Factor the given polynomial

y2βˆ’yrβˆ’30r2=

64

Factor the given polynomial

r2+ryβˆ’12y2=

65

Factor the given polynomial

r2βˆ’6rx+8x2=

66

Factor the given polynomial

t2βˆ’8tr+7r2=

67

Factor the given polynomial

t2+12ty+36y2=

68

Factor the given polynomial

x2+22xt+121t2=

69

Factor the given polynomial

x2βˆ’8xr+16r2=

70

Factor the given polynomial

x2βˆ’18xy+81y2=

71

Factor the given polynomial

8y2+24y+16=

72

Factor the given polynomial

3y2+18y+15=

73

Factor the given polynomial

2x2y+18xy+28y=

74

Factor the given polynomial

2x2y+14xy+24y=

75

Factor the given polynomial

2a2bβˆ’2abβˆ’12b=

76

Factor the given polynomial

4a2b+4abβˆ’8b=

77

Factor the given polynomial

4x2yβˆ’24xy+20y=

78

Factor the given polynomial

2x2yβˆ’10xy+12y=

79

Factor the given polynomial

10x3y+30x2y+20xy=

80

Factor the given polynomial

7x3y+21x2y+14xy=

81

Factor the given polynomial

x2y2βˆ’2x2yzβˆ’15x2z2=

82

Factor the given polynomial

x2y2+8x2yzβˆ’20x2z2=

83

Factor the given polynomial

(a+b)r2+10(a+b)r+16(a+b)=

84

Factor the given polynomial

(a+b)t2+11(a+b)t+30(a+b)=

85

Factor the given polynomial

t2+1.1t+0.18=

86

Factor the given polynomial

x2+1.1x+0.28=

87

Factor the given polynomial

x2y2+9xy+8=

88

Factor the given polynomial

x2y2+11xy+10=

89

Factor the given polynomial

y2r2βˆ’6yrβˆ’7=

90

Factor the given polynomial

y2x2βˆ’5yxβˆ’14=

91

Factor the given polynomial

r2t2βˆ’10rt+24=

92

Factor the given polynomial

r2t2βˆ’10rt+9=

93

Factor the given polynomial

2t2y2+8ty+6=

94

Factor the given polynomial

2t2r2+10tr+8=

95

Factor the given polynomial

6x2r2βˆ’18xrβˆ’24=

96

Factor the given polynomial

8x2r2+16xrβˆ’24=

97

Factor the given polynomial

2x2y3βˆ’16xy2+24y=

98

Factor the given polynomial

2x2y3βˆ’16xy2+24y=