Section7.4Factoring Trinomials with a Nontrivial Leading Coefficient
ΒΆIn 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.
Subsection7.4.1The AC Method
The AC Method is a technique for factoring trinomials like 4x2+5xβ6, 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 βwhat two numbers multiply to be β6 and add to be 5?β, we might come to the erroneous conclusion that 4x2+5xβ6 factors as (x+6)(xβ1). If we expand (x+6)(xβ1), we get
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+5xβ6.
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:
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.
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 βAC method,β 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:
Algorithm7.4.3The AC Method
To factor ax2+bx+c:
Multiply aβ c.
Make a table of factor pairs for ac. Look for a pair that adds to b. If you cannot find one, the polynomial is irreducible.
If you did find a factor pair summing to b, replace b with an explicit sum, and distribute x. With the four terms you have at this point, use factoring by grouping to continue. You are guaranteed to find a factorization.
Example7.4.4
Factor 10x2+23x+6.
10β 6=60
Use a list of factor pairs for 60 to find that 3 and 20 are a pair that sums to 23.
-
Intentionally break up the 23 as 3+20:
10x2β+23x+6=10x2β+3x+20x+6=(10x2+3x)+(20x+6)=x(10x+3)+2(10x+3)=(10x+3)(x+2)
Example7.4.5
Factor 2x2β5xβ3.
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.
Multiply \(2\cdot(-3)=-6\text{.}\)
-
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) -
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:
\(2x\) | \(1\) | |
\(x\) | \(2x^2\) | \(x\) |
\(-3\) | \(-6x\) | \(-3\) |
Our factorization passes the tests.
Example7.4.6
Factor 6p2+5pqβ6q2. Note that this example has two variables, but that does not really change our approach.
There is no greatest common factor. Since this is a trinomial, we try the AC Method.
Multiply \(6\cdot(-6)=-36\text{.}\)
-
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) -
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:
\(3p\) | \(-2q\) | |
\(2p\) | \(6p^2\) | \(-4pq\) |
\(3q\) | \(9pq\) | \(-6q^2\) |
Our factorization passes the tests.
Subsection7.4.2Factoring in Stages
Sometimes factoring a polynomial will take two or more βstages.β 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 18n2β21nβ60.
Notice that \(3\) is a common factor in this trinomial. We should factor it out first:
Now we are left with two factors, one of which is \(6n^2-7n-20\text{,}\) which might factor further. Using the AC Method:
\(6\cdot-20=-120\)
-
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) -
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 β16x3yβ12x2y+18xy.
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:
Now we are left with one factor being \(8x^2+6x-9\text{,}\) which might factor further. Using the AC Method:
\(8\cdot-9=-72\)
-
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) -
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+4rβ20=
4
Factor the given polynomial
2t2β19tβ10=
5
Factor the given polynomial
3t2β16t+20=
6
Factor the given polynomial
2t2β13t+21=
7
Factor the given polynomial
2x2β4x+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
6r2β5rβ14=
12
Factor the given polynomial
6r2β23rβ4=
13
Factor the given polynomial
8t2β23t+14=
14
Factor the given polynomial
4t2β7t+3=
15
Factor the given polynomial
10t2+9t+2=
16
Factor the given polynomial
4x2+16x+7=
17
Factor the given polynomial
12x2β7xβ10=
18
Factor the given polynomial
8y2+10yβ25=
19
Factor the given polynomial
15y2β23y+4=
20
Factor the given polynomial
6r2β19r+15=
21
Factor the given polynomial
12r2+18r+6=
22
Factor the given polynomial
35t2+42t+7=
23
Factor the given polynomial
45t2+27tβ18=
24
Factor the given polynomial
6t2+10tβ24=
25
Factor the given polynomial
4x2β14x+12=
26
Factor the given polynomial
10x2β12x+2=
27
Factor the given polynomial
15y4+24y3+9y2=
28
Factor the given polynomial
18y5+24y4+6y3=
29
Factor the given polynomial
6r6+28r5β10r4=
30
Factor the given polynomial
10r9+12r8β16r7=
31
Factor the given polynomial
6t9β28t8+16t7=
32
Factor the given polynomial
12t10β28t9+16t8=
33
Factor the given polynomial
2t2x2+15tx+18=
34
Factor the given polynomial
3x2r2+8xr+4=
35
Factor the given polynomial
3x2y2β5xyβ12=
36
Factor the given polynomial
5y2x2+2yxβ3=
37
Factor the given polynomial
2y2r2β9yr+4=
38
Factor the given polynomial
5r2y2β8ry+3=
39
Factor the given polynomial
3r2+14rt+15t2=
40
Factor the given polynomial
5t2+24tr+16r2=
41
Factor the given polynomial
3t2+5tyβ12y2=
42
Factor the given polynomial
3t2β19txβ14x2=
43
Factor the given polynomial
2x2β17xr+21r2=
44
Factor the given polynomial
3x2β16xr+21r2=
45
Factor the given polynomial
4y2+19yt+12t2=
46
Factor the given polynomial
4y2+15yr+9r2=
47
Factor the given polynomial
6r2β23rxβ4x2=
48
Factor the given polynomial
6r2βrtβ5t2=
49
Factor the given polynomial
8t2β23ty+14y2=
50
Factor the given polynomial
4t2β19tx+21x2=
51
Factor the given polynomial
25t2+30tr+8r2=
52
Factor the given polynomial
6x2+17xy+7y2=
53
Factor the given polynomial
10x2β9xyβ9y2=
54
Factor the given polynomial
6y2βytβ35t2=
55
Factor the given polynomial
25y2β25yx+4x2=
56
Factor the given polynomial
15r2β19rx+6x2=
57
Factor the given polynomial
15r2x2+18rx+3=
58
Factor the given polynomial
10t2y2+15ty+5=
59
Factor the given polynomial
24t2x2+8txβ16=
60
Factor the given polynomial
6t2r2+10trβ4=
61
Factor the given polynomial
12x6y2β28x5y+8x4=
62
Factor the given polynomial
12x8t2β20x7t+8x6=
63
Factor the given polynomial
12x2+28xy+8y2=
64
Factor the given polynomial
20x2+24xy+4y2=
65
Factor the given polynomial
9a2β21abβ18b2=
66
Factor the given polynomial
6a2β9abβ27b2=
67
Factor the given polynomial
4x2β10xy+4y2=
68
Factor the given polynomial
6x2β26xy+24y2=
69
Factor the given polynomial
4x2y+22xy2+28y3=
70
Factor the given polynomial
18x2y+27xy2+9y3=
71
Factor the given polynomial
8x2(yβ1)+20x(yβ1)+12(yβ1)=
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
-
Factor the given polynomial
5x2+22x+8=
-
Use your previous answer to factor
5(yβ8)2+22(yβ8)+8=
76
-
Factor the given polynomial
5x2+7x+2=
-
Use your previous answer to factor
5(y+3)2+7(y+3)+2=