Section7.4Factoring Trinomials with a Nontrivial Leading Coefficient
Â¶In SectionÂ 7.3, we learned how to factor \(ax^2+bx+c\) when \(a=1\text{.}\) In this section, we will examine the situation when \(a\neq1\text{.}\) The techniques are similar to those in the last section, but there are a few important differences that will makeorbreak your success in factoring these.
Subsection7.4.1The AC Method
The AC Method is a technique for factoring trinomials like \(4x^2+5x6\text{,}\) where there is no greatest common factor, and the leading coefficient is not \(1\text{.}\)
Please note at this point that if we try the method in the previous section and ask ourselves the question âwhat two numbers multiply to be \(6\) and add to be \(5\text{?}\)â, we might come to the erroneous conclusion that \(4x^2+5x6\) factors as \((x+6)(x1)\text{.}\) If we expand \((x+6)(x1)\text{,}\) we get
\begin{equation*} (x+6)(x1)=x^2+5x6 \end{equation*}This expression is almost correct, except for the missing leading coefficient, \(4\text{.}\) 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 AlgorithmÂ 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 \(4x^2+5x6\text{.}\)
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{equationacmethod}\tag{7.4.1} \end{align}When you encounter a trinomial like \(4x^2+5x6\) and you wish to factor it, the leading coefficient, \(4\text{,}\) is the \((pr)\) from EquationÂ (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 EquationÂ (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 EquationÂ (7.4.1), \((qr+ps)\text{.}\)
Can we do all this with the example \(4x^2+5x6\text{?}\) 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 \(4x^2+5x6\text{,}\) which is equal to the abstract \((qr+ps)\) from EquationÂ (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 SectionÂ 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+5x6}\amp=\left(4x^23x\right)+\left(8x6\right)\\ \amp=x\highlight{(4x3)}+2\highlight{(4x3)}\\ \amp=\highlight{(4x3)}(x+2) \end{align*}And this is the factorization of \(4x^2+5x6\text{.}\) This whole process is known as the âAC method,â 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 \(ax^2+bx+c\text{:}\)
Multiply \(a\cdot c\text{.}\)
Make a table of factor pairs for \(ac\text{.}\) Look for a pair that adds to \(b\text{.}\) If you cannot find one, the polynomial is irreducible.
If you did find a factor pair summing to \(b\text{,}\) replace \(b\) with an explicit sum, and distribute \(x\text{.}\) 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 \(10x^2+23x+6\text{.}\)
\(10\cdot6=60\)
Use a list of factor pairs for \(60\) to find that \(3\) and \(20\) are a pair that sums to \(23\text{.}\)

Intentionally break up the \(23\) as \(3+20\text{:}\)
\begin{align*} \amp10x^2\overbrace{{}+23x}+6\\ \amp=10x^2\overbrace{{}+3x+20x}+6\\ \amp=\left(10x^2+3x\right)+(20x+6)\\ \amp=x\highlight{(10x+3)}+2\highlight{(10x+3)}\\ \amp=\highlight{(10x+3)}(x+2) \end{align*}
Example7.4.5
Factor \(2x^25x3\text{.}\)
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 \(1\cdot6\) \(5\) (what we wanted) \(2\cdot3\) (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{{}+x6x}3\\ \amp=\left(2x^2+x\right)+(6x3)\\ \amp=x\highlight{(2x+1)}3\highlight{(2x+1)}\\ \amp=\highlight{(2x+1)}(x3) \end{align*}
So we believe that \(2x^25x3\) factors as \((2x+1)(x3)\text{,}\) 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 \(6p^2+5pq6q^2\text{.}\) 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 \(1\cdot36\) \(35\) \(2\cdot18\) \(16\) \(3\cdot12\) \(9\) \(4\cdot9\) \(5\) (close; wrong sign) \(6\cdot6\) \(0\) 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^24pq\right)+\left(9pq6q^2\right)\\ \amp=2p\highlight{(3p2q)}+3q\highlight{(3p2q)}\\ \amp=\highlight{(3p2q)}(2p+3q) \end{align*}
So we believe that \(6p^2+5pq6q^2\) factors as \((3p2q)(2p+3q)\text{,}\) 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 âstages.â 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 \(18n^221n60\text{.}\)
Notice that \(3\) is a common factor in this trinomial. We should factor it out first:
\begin{equation*} 18n^221n60=3\left(6n^27n20\right) \end{equation*}Now we are left with two factors, one of which is \(6n^27n20\text{,}\) which might factor further. Using the AC Method:
\(6\cdot20=120\)

Examine factor pairs that multiply to \(120\text{,}\) looking for a pair that sums to \(7\text{:}\)
Factor Pair Sum of the Pair \(1\cdot120\) \(119\) \(2\cdot60\) \(58\) \(3\cdot40\) \(37\) \(4\cdot30\) \(26\) \(5\cdot24\) \(19\) \(6\cdot20\) \(14\) \(8\cdot15\) \(7\) (what we wanted) \(10\cdot12\) (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^221n60 \amp=3\left(6n^2\overbrace{{}7n}20\right)\\ \amp=3\left(6n^2\overbrace{{}+8n15n}20\right)\\ \amp=3\left((6n^2+8n)+(15n20)\right)\\ \amp=3\left(2n\highlight{(3n+4)}5\highlight{(3n+4)}\right)\\ \amp=3\highlight{(3n+4)}(2n5) \end{align*}
So we believe that \(18n^221n60\) factors as \(3(3n+4)(2n5)\text{,}\) and you should check by multiplying out the factored form.
Example7.4.8
Factor \(16x^3y12x^2y+18xy\text{.}\)
Notice that \(2xy\) is a common factor in this trinomial. Also the leading coefficient is negative, and as discussed in SectionÂ 7.1, it is wise to factor that out as well. So we find:
\begin{equation*} 16x^3y12x^2y+18xy=2xy\left(8x^2+6x9\right) \end{equation*}Now we are left with one factor being \(8x^2+6x9\text{,}\) which might factor further. Using the AC Method:
\(8\cdot9=72\)

Examine factor pairs that multiply to \(72\text{,}\) looking for a pair that sums to \(6\text{:}\)
Factor Pair Sum of the Pair \(1\cdot72\) \(71\) \(2\cdot36\) \(34\) \(3\cdot24\) \(21\) \(4\cdot18\) \(14\) \(6\cdot12\) \(6\) (close; wrong sign) \(8\cdot9\) \(1\) 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^3y12x^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^26x)+(12x9)\right)\\ \amp=2xy\left(2x\highlight{(4x3)}+3\highlight{(4x3)}\right)\\ \amp=2xy\highlight{(4x3)}(2x+3) \end{align*}
So we believe that \(16x^3y12x^2y+18xy\) factors as \(2xy(4x3)(2x+3)\text{,}\) and you should check by multiplying out the factored form.
Subsection7.4.3Exercises
1
Factor the given polynomial
\({3y^{2}+10y+8}=\)
2
Factor the given polynomial
\({2r^{2}+7r+6}=\)
3
Factor the given polynomial
\({5r^{2}8r4}=\)
4
Factor the given polynomial
\({3t^{2}13t10}=\)
5
Factor the given polynomial
\({2t^{2}17t+8}=\)
6
Factor the given polynomial
\({2x^{2}11x+5}=\)
7
Factor the given polynomial
\({3x^{2}+6x+5}=\)
8
Factor the given polynomial
\({2x^{2}+x+2}=\)
9
Factor the given polynomial
\({6y^{2}+19y+14}=\)
10
Factor the given polynomial
\({4y^{2}+19y+12}=\)
11
Factor the given polynomial
\({4r^{2}+r5}=\)
12
Factor the given polynomial
\({4r^{2}+r5}=\)
13
Factor the given polynomial
\({4t^{2}23t+15}=\)
14
Factor the given polynomial
\({8t^{2}27t+9}=\)
15
Factor the given polynomial
\({8x^{2}+18x+7}=\)
16
Factor the given polynomial
\({8x^{2}+10x+3}=\)
17
Factor the given polynomial
\({20x^{2}+x12}=\)
18
Factor the given polynomial
\({25y^{2}25y14}=\)
19
Factor the given polynomial
\({20y^{2}19y+3}=\)
20
Factor the given polynomial
\({15r^{2}29r+8}=\)
21
Factor the given polynomial
\({25r^{2}+35r+10}=\)
22
Factor the given polynomial
\({10t^{2}+26t+16}=\)
23
Factor the given polynomial
\({20t^{2}+30t20}=\)
24
Factor the given polynomial
\({18t^{2}+6t12}=\)
25
Factor the given polynomial
\({4x^{2}18x+14}=\)
26
Factor the given polynomial
\({15x^{2}20x+5}=\)
27
Factor the given polynomial
\({15y^{8}+24y^{7}+9y^{6}}=\)
28
Factor the given polynomial
\({9y^{7}+30y^{6}+9y^{5}}=\)
29
Factor the given polynomial
\({4r^{5}6r^{4}10r^{3}}=\)
30
Factor the given polynomial
\({9r^{6}15r^{5}6r^{4}}=\)
31
Factor the given polynomial
\({9t^{8}15t^{7}+6t^{6}}=\)
32
Factor the given polynomial
\({6t^{8}26t^{7}+20t^{6}}=\)
33
Factor the given polynomial
\({3t^{2}y^{2}+8ty+4}=\)
34
Factor the given polynomial
\({5x^{2}t^{2}+11xt+2}=\)
35
Factor the given polynomial
\({2x^{2}r^{2}xr10}=\)
36
Factor the given polynomial
\({5y^{2}t^{2}7yt24}=\)
37
Factor the given polynomial
\({2y^{2}t^{2}17yt+21}=\)
38
Factor the given polynomial
\({5r^{2}x^{2}18rx+16}=\)
39
Factor the given polynomial
\({2r^{2}+17rx+8x^{2}}=\)
40
Factor the given polynomial
\({5t^{2}+12tr+4r^{2}}=\)
41
Factor the given polynomial
\({5t^{2}16tr16r^{2}}=\)
42
Factor the given polynomial
\({3t^{2}26tx9x^{2}}=\)
43
Factor the given polynomial
\({5x^{2}23xt+12t^{2}}=\)
44
Factor the given polynomial
\({2x^{2}19xy+9y^{2}}=\)
45
Factor the given polynomial
\({4y^{2}+9yx+5x^{2}}=\)
46
Factor the given polynomial
\({4y^{2}+17yt+18t^{2}}=\)
47
Factor the given polynomial
\({4r^{2}17ry15y^{2}}=\)
48
Factor the given polynomial
\({8r^{2}15rx2x^{2}}=\)
49
Factor the given polynomial
\({4t^{2}19ty+12y^{2}}=\)
50
Factor the given polynomial
\({8t^{2}21ty+10y^{2}}=\)
51
Factor the given polynomial
\({8t^{2}+14tx+3x^{2}}=\)
52
Factor the given polynomial
\({10x^{2}+27xr+5r^{2}}=\)
53
Factor the given polynomial
\({25x^{2}25xy6y^{2}}=\)
54
Factor the given polynomial
\({10y^{2}+yx3x^{2}}=\)
55
Factor the given polynomial
\({12y^{2}19yr+5r^{2}}=\)
56
Factor the given polynomial
\({10r^{2}27ry+5y^{2}}=\)
57
Factor the given polynomial
\({10r^{2}t^{2}+25rt+10}=\)
58
Factor the given polynomial
\({30t^{2}r^{2}+40tr+10}=\)
59
Factor the given polynomial
\({15t^{2}y^{2}+25ty10}=\)
60
Factor the given polynomial
\({4t^{2}x^{2}+2tx12}=\)
61
Factor the given polynomial
\({4x^{8}r^{2}18x^{7}r+8x^{6}}=\)
62
Factor the given polynomial
\({25x^{9}r^{2}30x^{8}r+5x^{7}}=\)
63
Factor the given polynomial
\({4x^{2}+30xy+14y^{2}}=\)
64
Factor the given polynomial
\({15x^{2}+20xy+5y^{2}}=\)
65
Factor the given polynomial
\({9a^{2}+3ab12b^{2}}=\)
66
Factor the given polynomial
\({45a^{2}+9ab36b^{2}}=\)
67
Factor the given polynomial
\({4x^{2}18xy+8y^{2}}=\)
68
Factor the given polynomial
\({10x^{2}25xy+15y^{2}}=\)
69
Factor the given polynomial
\({4x^{2}y+22xy^{2}+10y^{3}}=\)
70
Factor the given polynomial
\({8x^{2}y+28xy^{2}+20y^{3}}=\)
71
Factor the given polynomial
\({4x^{2}\!\left(y3\right)+22x\!\left(y3\right)+10\!\left(y3\right)}=\)
72
Factor the given polynomial
\({12x^{2}\!\left(y4\right)+32x\!\left(y4\right)+16\!\left(y4\right)}=\)
73
Factor the given polynomial
\({6x^{2}\!\left(y+2\right)+9x\!\left(y+2\right)+3\!\left(y+2\right)}=\)
74
Factor the given polynomial
\({24x^{2}\!\left(y6\right)+32x\!\left(y6\right)+8\!\left(y6\right)}=\)
75

Factor the given polynomial
\({3x^{2}+14x+16}=\)

Use your previous answer to factor
\({3\!\left(y+8\right)^{2}+14\!\left(y+8\right)+16}=\)
76

Factor the given polynomial
\({2x^{2}+13x+18}=\)

Use your previous answer to factor
\({2\!\left(y+7\right)^{2}+13\!\left(y+7\right)+18}=\)