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Section7.4Factoring Trinomials with a Nontrivial Leading Coefficient

ObjectivesPCC Course Content and Outcome Guide

In Section 7.3, we learned how to factor ax2+bx+c when a=1. In this section, we will examine the situation when aβ‰ 1. The techniques are similar to those in the last section, but there are a few important differences that will make-or-break your success in factoring these.

Figure7.4.1Alternative Video Lesson

Subsection7.4.1The AC Method

The AC Method is a technique for factoring trinomials like 4x2+5xβˆ’6, where there is no greatest common factor, and the leading coefficient is not 1.

Please note at this point that if we try the method in the previous section and ask ourselves the question β€œwhat two numbers multiply to be βˆ’6 and add to be 5?”, we might come to the erroneous conclusion that 4x2+5xβˆ’6 factors as (x+6)(xβˆ’1). If we expand (x+6)(xβˆ’1), we get

(x+6)(xβˆ’1)=x2+5xβˆ’6

This expression is almost correct, except for the missing leading coefficient, 4. Dealing with this missing coefficient requires starting over with the AC method. If you are only interested in the steps for using the technique, skip ahead to Algorithm 7.4.3.

The example below explains why the AC Method works. Understanding all of the details might take a few rereads, and coming back to this example after mastering the algorithm may be the best course of action.

Example7.4.2

Expand the expression (px+q)(rx+s) and analyze the result to gain an insight into why the AC method works. Then use this information to factor 4x2+5xβˆ’6.

Solution

Factoring is the opposite process from multiplying polynomials together. We can gain some insight into how to factor complicated polynomials by taking a closer look at what happens when two generic polynomials are multiplied together:

\begin{align} (px+q)(rx+s)\amp=(px+q)(rx)+(px+q)s\notag\\ \amp=(px)(rx)+q(rx)+(px)s+qs\notag\\ \amp=(pr)x^2+qrx+psx+qs\notag\\ \amp=(pr)x^2+(qr+ps)x+qs\label{equation-ac-method}\tag{7.4.1} \end{align}

When you encounter a trinomial like \(4x^2+5x-6\) and you wish to factor it, the leading coefficient, \(4\text{,}\) is the \((pr)\) from Equation (7.4.1). Similarly, the \(-6\) is the \(qs\text{,}\) and the \(5\) is the \((qr+ps)\text{.}\)

Now, if you multiply the leading coefficient and constant term from Equation (7.4.1), you have \((pr)(qs)\text{,}\) which equals \(pqrs\text{.}\) Notice that if we factor this number in just the right way, \((qr)(ps)\text{,}\) then we have two factors that add to the middle coefficient from Equation (7.4.1), \((qr+ps)\text{.}\)

Can we do all this with the example \(4x^2+5x-6\text{?}\) Multiplying \(4\) and \(-6\) makes \(-24\text{.}\) Is there some way to factor \(-24\) into two factors which add to \(5\text{?}\) We make a table of factor pairs for \(-24\) to see:

Factor Pair Sum of the Pair
\(-1\cdot24\) \(23\)
\(-2\cdot12\) \(10\)
\(-3\cdot8\) \(5\) (what we wanted)
\(-4\cdot6\) (no need to go this far)
Factor Pair Sum of the Pair
\(1\cdot(-24)\) (no need to go this far)
\(2\cdot(-12)\) (no need to go this far)
\(3\cdot(-8)\) (no need to go this far)
\(4\cdot(-6)\) (no need to go this far)

So that \(5\) in \(4x^2+5x-6\text{,}\) which is equal to the abstract \((qr+ps)\) from Equation (7.4.1), breaks down as \(-3+8\text{.}\) We can take \(-3\) to be the \(qr\) and \(8\) to be the \(ps\text{.}\) Once we intentionally break up the \(5\) this way, factoring by grouping (see Section 7.2) can take over and is guaranteed to give us a factorization.

\begin{align*} 4x^2\overbrace{{}+5x}-6\amp=4x^2\overbrace{-3x+8x}-6\\ \end{align*}

Now that there are four terms, group them and factor out each group's greatest common factor.

\begin{align*} \phantom{4x^2+5x-6}\amp=\left(4x^2-3x\right)+\left(8x-6\right)\\ \amp=x\highlight{(4x-3)}+2\highlight{(4x-3)}\\ \amp=\highlight{(4x-3)}(x+2) \end{align*}

And this is the factorization of \(4x^2+5x-6\text{.}\) This whole process is known as the β€œAC method,” since it begins by multiplying \(a\) and \(c\) from the generic \(ax^2+bx+c\text{.}\)

The AC Method

Here is a summary of the algorithm:

Example7.4.4

Factor 10x2+23x+6.

  1. 10β‹…6=60

  2. Use a list of factor pairs for 60 to find that 3 and 20 are a pair that sums to 23.

  3. Intentionally break up the 23 as 3+20:

    10x2⏞+23x+6=10x2⏞+3x+20x+6=(10x2+3x)+(20x+6)=x(10x+3)+2(10x+3)=(10x+3)(x+2)
Example7.4.5

Factor 2x2βˆ’5xβˆ’3.

Solution

Always start the factoring process by examining if there is a greatest common factor. Here there is not one. Next, note that this is a trinomial with a leading coefficient that is not \(1\text{.}\) So the AC Method may be of help.

  1. Multiply \(2\cdot(-3)=-6\text{.}\)

  2. Examine factor pairs that multiply to \(-6\text{,}\) looking for a pair that sums to \(-5\text{:}\)

    Factor Pair Sum of the Pair
    \(1\cdot-6\) \(-5\) (what we wanted)
    \(2\cdot-3\) (no need to go this far)
    Factor Pair Sum of the Pair
    \(-1\cdot6\) (no need to go this far)
    \(-2\cdot3\) (no need to go this far)
  3. Intentionally break up the \(-5\) as \(1+(-6)\text{:}\)

    \begin{align*} 2x^2\overbrace{{}-5x}-3\amp=2x^2\overbrace{{}+x-6x}-3\\ \amp=\left(2x^2+x\right)+(-6x-3)\\ \amp=x\highlight{(2x+1)}-3\highlight{(2x+1)}\\ \amp=\highlight{(2x+1)}(x-3) \end{align*}

So we believe that \(2x^2-5x-3\) factors as \((2x+1)(x-3)\text{,}\) and we should check by multiplying out the factored form:

\begin{align*} (2x+1)(x-3)\amp=(2x+1)\cdot x+(2x+1)\cdot(-3)\\ \amp=2x^2+x-6x-3\\ \amp\stackrel{\checkmark}{=}2x^2-5x-3 \end{align*}
\(2x\) \(1\)
\(x\) \(2x^2\) \(x\)
\(-3\) \(-6x\) \(-3\)

Our factorization passes the tests.

Example7.4.6

Factor 6p2+5pqβˆ’6q2. Note that this example has two variables, but that does not really change our approach.

Solution

There is no greatest common factor. Since this is a trinomial, we try the AC Method.

  1. Multiply \(6\cdot(-6)=-36\text{.}\)

  2. Examine factor pairs that multiply to \(-36\text{,}\) looking for a pair that sums to \(5\text{:}\)

    Factor Pair Sum of the Pair
    \(1\cdot-36\) \(-35\)
    \(2\cdot-18\) \(-16\)
    \(3\cdot-12\) \(-9\)
    \(4\cdot-9\) \(-5\) (close; wrong sign)
    \(6\cdot-6\) \(0\)
    Factor Pair Sum of the Pair
    \(-1\cdot36\) \(35\)
    \(-2\cdot18\) \(16\)
    \(-3\cdot12\) \(9\)
    \(-4\cdot9\) \(5\) (what we wanted)
  3. Intentionally break up the \(5\) as \(-4+9\text{:}\)

    \begin{align*} 6p^2\overbrace{{}+5pq}-6q^2\amp=6p^2\overbrace{{}-4pq+9pq}-6q^2\\ \amp=\left(6p^2-4pq\right)+\left(9pq-6q^2\right)\\ \amp=2p\highlight{(3p-2q)}+3q\highlight{(3p-2q)}\\ \amp=\highlight{(3p-2q)}(2p+3q) \end{align*}

So we believe that \(6p^2+5pq-6q^2\) factors as \((3p-2q)(2p+3q)\text{,}\) and we should check by multiplying out the factored form:

\begin{align*} (3p-2q)(2p+3q)\amp=(3p-2q)\cdot 2p+(3p-2q)\cdot3q\\ \amp=6p^2-4pq+9pq-6q^2\\ \amp\stackrel{\checkmark}{=}6p^2+5pq-6q^2 \end{align*}
\(3p\) \(-2q\)
\(2p\) \(6p^2\) \(-4pq\)
\(3q\) \(9pq\) \(-6q^2\)

Our factorization passes the tests.

Subsection7.4.2Factoring in Stages

Sometimes factoring a polynomial will take two or more β€œstages.” For instance you may need to 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.

Example7.4.7

Factor 18n2βˆ’21nβˆ’60.

Solution

Notice that \(3\) is a common factor in this trinomial. We should factor it out first:

\begin{equation*} 18n^2-21n-60=3\left(6n^2-7n-20\right) \end{equation*}

Now we are left with two factors, one of which is \(6n^2-7n-20\text{,}\) which might factor further. Using the AC Method:

  1. \(6\cdot-20=-120\)

  2. Examine factor pairs that multiply to \(-120\text{,}\) looking for a pair that sums to \(-7\text{:}\)

    Factor Pair Sum of the Pair
    \(1\cdot-120\) \(-119\)
    \(2\cdot-60\) \(-58\)
    \(3\cdot-40\) \(-37\)
    \(4\cdot-30\) \(-26\)
    \(5\cdot-24\) \(-19\)
    \(6\cdot-20\) \(-14\)
    \(8\cdot-15\) \(-7\) (what we wanted)
    \(10\cdot-12\) (no need to go this far)
    Factor Pair Sum of the Pair
    \(-1\cdot120\) (no need to go this far)
    \(-2\cdot60\) (no need to go this far)
    \(-3\cdot40\) (no need to go this far)
    \(-4\cdot30\) (no need to go this far)
    \(-5\cdot24\) (no need to go this far)
    \(-6\cdot20\) (no need to go this far)
    \(-8\cdot15\) (no need to go this far)
    \(-10\cdot12\) (no need to go this far)
  3. Intentionally break up the \(-7\) as \(8+(-15)\text{:}\)

    \begin{align*} 18n^2-21n-60 \amp=3\left(6n^2\overbrace{{}-7n}-20\right)\\ \amp=3\left(6n^2\overbrace{{}+8n-15n}-20\right)\\ \amp=3\left((6n^2+8n)+(-15n-20)\right)\\ \amp=3\left(2n\highlight{(3n+4)}-5\highlight{(3n+4)}\right)\\ \amp=3\highlight{(3n+4)}(2n-5) \end{align*}

So we believe that \(18n^2-21n-60\) factors as \(3(3n+4)(2n-5)\text{,}\) and you should check by multiplying out the factored form.

Example7.4.8

Factor βˆ’16x3yβˆ’12x2y+18xy.

Solution

Notice that \(2xy\) is a common factor in this trinomial. Also the leading coefficient is negative, and as discussed in Section 7.1, it is wise to factor that out as well. So we find:

\begin{equation*} -16x^3y-12x^2y+18xy=-2xy\left(8x^2+6x-9\right) \end{equation*}

Now we are left with one factor being \(8x^2+6x-9\text{,}\) which might factor further. Using the AC Method:

  1. \(8\cdot-9=-72\)

  2. Examine factor pairs that multiply to \(-72\text{,}\) looking for a pair that sums to \(6\text{:}\)

    Factor Pair Sum of the Pair
    \(1\cdot-72\) \(-71\)
    \(2\cdot-36\) \(-34\)
    \(3\cdot-24\) \(-21\)
    \(4\cdot-18\) \(-14\)
    \(6\cdot-12\) \(-6\) (close; wrong sign)
    \(8\cdot-9\) \(-1\)
    Factor Pair Sum of the Pair
    \(-1\cdot72\) \(71\)
    \(-2\cdot36\) \(34\)
    \(-3\cdot24\) \(21\)
    \(-4\cdot18\) \(14\)
    \(-6\cdot12\) \(6\) (what we wanted)
    \(-8\cdot9\) (no need to go this far)
  3. Intentionally break up the \(6\) as \(-6+12\text{:}\)

    \begin{align*} -16x^3y-12x^2y+18xy\amp=-2xy\left(8x^2\overbrace{{}+6x}-9\right)\\ \amp=-2xy\left(8x^2\overbrace{{}-6x+12x}-9\right)\\ \amp=-2xy\left((8x^2-6x)+(12x-9)\right)\\ \amp=-2xy\left(2x\highlight{(4x-3)}+3\highlight{(4x-3)}\right)\\ \amp=-2xy\highlight{(4x-3)}(2x+3) \end{align*}

So we believe that \(-16x^3y-12x^2y+18xy\) factors as \(-2xy(4x-3)(2x+3)\text{,}\) and you should check by multiplying out the factored form.

SubsectionExercises

1

Factor the given polynomial

3y2+11y+8=

2

Factor the given polynomial

5r2+13r+6=

3

Factor the given polynomial

3r2+4rβˆ’20=

4

Factor the given polynomial

2t2βˆ’19tβˆ’10=

5

Factor the given polynomial

3t2βˆ’16t+20=

6

Factor the given polynomial

2t2βˆ’13t+21=

7

Factor the given polynomial

2x2βˆ’4x+5=

8

Factor the given polynomial

2x2+7x+10=

9

Factor the given polynomial

6y2+23y+15=

10

Factor the given polynomial

4y2+13y+3=

11

Factor the given polynomial

6r2βˆ’5rβˆ’14=

12

Factor the given polynomial

6r2βˆ’23rβˆ’4=

13

Factor the given polynomial

8t2βˆ’23t+14=

14

Factor the given polynomial

4t2βˆ’7t+3=

15

Factor the given polynomial

10t2+9t+2=

16

Factor the given polynomial

4x2+16x+7=

17

Factor the given polynomial

12x2βˆ’7xβˆ’10=

18

Factor the given polynomial

8y2+10yβˆ’25=

19

Factor the given polynomial

15y2βˆ’23y+4=

20

Factor the given polynomial

6r2βˆ’19r+15=

21

Factor the given polynomial

12r2+18r+6=

22

Factor the given polynomial

35t2+42t+7=

23

Factor the given polynomial

45t2+27tβˆ’18=

24

Factor the given polynomial

6t2+10tβˆ’24=

25

Factor the given polynomial

4x2βˆ’14x+12=

26

Factor the given polynomial

10x2βˆ’12x+2=

27

Factor the given polynomial

15y4+24y3+9y2=

28

Factor the given polynomial

18y5+24y4+6y3=

29

Factor the given polynomial

6r6+28r5βˆ’10r4=

30

Factor the given polynomial

10r9+12r8βˆ’16r7=

31

Factor the given polynomial

6t9βˆ’28t8+16t7=

32

Factor the given polynomial

12t10βˆ’28t9+16t8=

33

Factor the given polynomial

2t2x2+15tx+18=

34

Factor the given polynomial

3x2r2+8xr+4=

35

Factor the given polynomial

3x2y2βˆ’5xyβˆ’12=

36

Factor the given polynomial

5y2x2+2yxβˆ’3=

37

Factor the given polynomial

2y2r2βˆ’9yr+4=

38

Factor the given polynomial

5r2y2βˆ’8ry+3=

39

Factor the given polynomial

3r2+14rt+15t2=

40

Factor the given polynomial

5t2+24tr+16r2=

41

Factor the given polynomial

3t2+5tyβˆ’12y2=

42

Factor the given polynomial

3t2βˆ’19txβˆ’14x2=

43

Factor the given polynomial

2x2βˆ’17xr+21r2=

44

Factor the given polynomial

3x2βˆ’16xr+21r2=

45

Factor the given polynomial

4y2+19yt+12t2=

46

Factor the given polynomial

4y2+15yr+9r2=

47

Factor the given polynomial

6r2βˆ’23rxβˆ’4x2=

48

Factor the given polynomial

6r2βˆ’rtβˆ’5t2=

49

Factor the given polynomial

8t2βˆ’23ty+14y2=

50

Factor the given polynomial

4t2βˆ’19tx+21x2=

51

Factor the given polynomial

25t2+30tr+8r2=

52

Factor the given polynomial

6x2+17xy+7y2=

53

Factor the given polynomial

10x2βˆ’9xyβˆ’9y2=

54

Factor the given polynomial

6y2βˆ’ytβˆ’35t2=

55

Factor the given polynomial

25y2βˆ’25yx+4x2=

56

Factor the given polynomial

15r2βˆ’19rx+6x2=

57

Factor the given polynomial

15r2x2+18rx+3=

58

Factor the given polynomial

10t2y2+15ty+5=

59

Factor the given polynomial

24t2x2+8txβˆ’16=

60

Factor the given polynomial

6t2r2+10trβˆ’4=

61

Factor the given polynomial

12x6y2βˆ’28x5y+8x4=

62

Factor the given polynomial

12x8t2βˆ’20x7t+8x6=

63

Factor the given polynomial

12x2+28xy+8y2=

64

Factor the given polynomial

20x2+24xy+4y2=

65

Factor the given polynomial

9a2βˆ’21abβˆ’18b2=

66

Factor the given polynomial

6a2βˆ’9abβˆ’27b2=

67

Factor the given polynomial

4x2βˆ’10xy+4y2=

68

Factor the given polynomial

6x2βˆ’26xy+24y2=

69

Factor the given polynomial

4x2y+22xy2+28y3=

70

Factor the given polynomial

18x2y+27xy2+9y3=

71

Factor the given polynomial

8x2(yβˆ’1)+20x(yβˆ’1)+12(yβˆ’1)=

72

Factor the given polynomial

4x2(y+8)+30x(y+8)+14(y+8)=

73

Factor the given polynomial

6x2(y+9)+26x(y+9)+20(y+9)=

74

Factor the given polynomial

10x2(y+4)+32x(y+4)+6(y+4)=

75
  1. Factor the given polynomial

    5x2+22x+8=

  2. Use your previous answer to factor

    5(yβˆ’8)2+22(yβˆ’8)+8=

76
  1. Factor the given polynomial

    5x2+7x+2=

  2. Use your previous answer to factor

    5(y+3)2+7(y+3)+2=