In this section, you will....
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learn the \(ac\)-method of factoring
Section 4.4 Factoring Trinomials Using the \(ac\)-method
In the previous section, we looked at three factoring techniques. In this section, we will learn a fourth factoring technique that allows up to factor a type of polynomial called a trinomial.
Supplemental Videos.
The main topics of this section are also presented in the following videos:
Some trinomials of the form \(ax^2+bx+c\) can be factored as a product of binomials using a technique called the \(ac\)-method. If such a trinomial can be factored, then the middle term, \(bx\text{,}\) can be replaced with two terms with coefficients whose sum is \(b\) and whose product is \(ac\text{.}\) This substitution results in an equivalent expression with four terms that can be factored by grouping.
If factors of \(ac\) cannot be found to add up to \(b\) then the trinomial is prime.
Example 4.35.
Factor \(x^2+12x+20\text{.}\)
Solution.
Here, \(a=1\text{,}\) \(b=12\text{,}\) and \(c=20\text{,}\) so we need to find two numbers whose sum is \(12\) and whose product is \(20\text{.}\) To do this, list all factorizations of \(20\) and search for factors whose sum equals \(12\text{.}\)
| \(1\cdot 20\) | \(1 +20=21\) |
| \(2\cdot 10\) | \(\alert{2 +10=12}\) |
| \(4\cdot 5\) | \(4 +5=9\) |
Choose \(20 = 2 \cdot 10\) because \(2 + 10 = 12\text{.}\) Therefore, \(12x=2x+10x\text{,}\) and we can write
\begin{equation*}
x^2+\alert{12x}+20=x^2+\alert{2x+10x}+20.
\end{equation*}
We can now factor the equivalent expression by grouping.
\begin{equation*}
\begin{aligned} x^2+12x+20\amp = x^2+2x+10x+20\\ \amp = x(x+2)+10(x+2)\\ \amp = (x+10)(x+2) \end{aligned}
\end{equation*}
Our factored form is \((x+10)(x+2)\text{.}\)
We check by multiplying the two binomials:
\begin{equation*}
\begin{aligned} (x+2)(x+10)\amp =x^2+10x+2x+20\\ \amp = x^2+12x+20 \end{aligned}
\end{equation*}
Note: We made a choice in the above problem to write \(12x=2x+10x\text{.}\) What would happen if we instead had written \(12x=10x+2x\text{?}\) Notice that in the following calculation the answer is equivalent.
\begin{equation*}
x^2+\alert{12x}+20=x^2+\alert{10x+2x}+20.
\end{equation*}
We factor the equivalent expression by grouping.
\begin{equation*}
\begin{aligned} x^2+\alert{12x}+20 \amp= x^2+\alert{10x+2x}+20\\ \amp = x(x+10)+2(x+10)\\ \amp = (x+2)(x+10)\\ \amp = (x+10)(x+2) \end{aligned}
\end{equation*}
Since multiplication is commutative, the order of the factors does not matter.
\begin{equation*}
\begin{aligned} x^2+12x+20 \amp = (x+2)(x+10)\\ \amp = (x+10)(x+2) \end{aligned}
\end{equation*}
If the last term of the trinomial is positive, then either both of the constant factors must be negative or both must be positive.
Example 4.37.
Factor \(x^2y^2-7xy+12\text{.}\)
Solution.
Even though there are multiple variables, this technique will still work. Here, \(a=1\text{,}\) \(b=-7\text{,}\) and \(c=12\text{,}\) so we need to find two numbers whose sum is \(-7\) and whose product is \(12\text{.}\) Notice that in this case, both numbers must be negative.
| \(-1\cdot -12\) | \(-1 -12=-13\) |
| \(-2\cdot -6\) | \(-2-6=-8\) |
| \(-3\cdot -4\) | \(\alert{-3-4=-7}\) |
Choose \(12 = -3 \cdot -4\) because \(-3 -4 = -7\text{.}\) Therefore, \(-7xy=-3xy-4xy\text{,}\) and we can write
\begin{equation*}
x^2y^2\alert{-7xy}+12=x^2y^2\alert{-3xy-4xy}+12.
\end{equation*}
We factor the equivalent expression by grouping.
\begin{equation*}
\begin{aligned} x^2y^2-7xy+12\amp = x^2y^2-3xy-4xy+12\\ \amp = xy(xy-3)-4(xy-3)\\ \amp = (xy-4)(xy-3) \end{aligned}
\end{equation*}
Our factored form is \((xy-4)(xy-3)\text{.}\)
We check by multiplying the two binomials.
\begin{equation*}
\begin{aligned} (xy-4)(xy-3)\amp =(xy)^2-3xy-4xy+12\\ \amp = x^2y^2-7xy+12 \end{aligned}
\end{equation*}
The check is very important and is something you should plan to do on every single problem.
Keep in mind that some polynomials are what is called prime. For example, consider the trinomial \(x^2+3x+20\) and the factors of 20:
\begin{equation*}
\begin{aligned} 20\amp =1\cdot 20\\ \amp = 2\cdot 10\\ \amp=4\cdot 5 \end{aligned}
\end{equation*}
There are no integer factors of 20 whose sum is 3. Therefore, the original trinomial cannot be factored as a product of two binomials with integer coefficients. The trinomial is prime, that is, it cannot be factored with integer terms.
In all of the examples we’ve seen so far, the leading coefficient has been 1. We will now see what happens when the leading coefficient is not 1.
Example 4.39.
Factor \(18x^2-31x+6\) using the \(ac\)-method.
Solution.
Here \(a=18,\, b=-31\text{,}\) and \(c=6\text{.}\)
\begin{equation*}
\begin{aligned} ac=18(6)=108 \end{aligned}
\end{equation*}
Now we factor \(108\text{,}\) and search for factors whose sum is \(-31\text{.}\)
\begin{equation*}
\begin{aligned} 108\amp = -1(-108)\\ \amp = -2(-54)\\ \amp= -3(-36)\\ \amp = \alert{-4(-27)}\\ \amp = -6(-18)\\ \amp = -9(-12) \end{aligned}
\end{equation*}
In this case, the sum of the factors \(-27\) and \(-4\) equals the middle coefficient, \(-31\text{.}\) Therefore, \(-31x=-27x-4x\text{,}\) and we can write
\begin{equation*}
\begin{aligned} 18x^2\alert{-31x}+6=18x^2\alert{-27x-4x}+6\text{.} \end{aligned}
\end{equation*}
We factor the equivalent expression by grouping.
\begin{equation*}
\begin{aligned} 18x^2-31x+6\amp = 18x^2-27x-4x+6\\ \amp = 9x(2x-3)-2(2x-3)\\ \amp = (2x-3)(9x-2) \end{aligned}
\end{equation*}
Our factored form is \((2x-3)(9x-2)\text{.}\)
Example 4.40.
Factor \(4x^2y^2-7xy-15\) using the \(ac\)-method.
Solution.
Here \(a=4,\, b=-7\text{,}\) and \(c=-15\text{.}\)
\begin{equation*}
\begin{aligned} ac=4(-15)=-60 \end{aligned}
\end{equation*}
Now we factor \(-60\text{,}\) and search for factors whose sum is \(-7\text{.}\)
\begin{equation*}
\begin{aligned} -60\amp = 1(-60)\\ \amp = 2(-30)\\ \amp= 3(-20)\\ \amp = 4(-15)\\ \amp = \alert{5(-12)}\\ \amp = 6(-10) \end{aligned}
\end{equation*}
In this case, the sum of the factors \(5\) and \(-12\) equals the middle coefficient, \(-7\text{.}\) Therefore, we replace \(-7xy\) with \(5xy-12xy\text{.}\)
\begin{equation*}
\begin{aligned} 4x^2y^2-7xy-15\amp = 4x^2y^2+5xy-12xy-15\\ \amp =xy(4xy+5)-3(4xy+5)\\ \amp = (4xy+5)(xy-3) \end{aligned}
\end{equation*}
Our factored form is \((4xy+5)(xy-3)\text{.}\)
Example 4.41.
Factor \(5x^2+16xy+3y^2\text{.}\)
Solution.
Here \(a=5,\, b=16y\text{,}\) and \(c=3y^2\text{.}\)
\begin{equation*}
\begin{aligned} ac=5(3y^2)=15y^2 \end{aligned}
\end{equation*}
Now we factor \(15y^2\text{,}\) and search for factors whose sum is \(16y\text{.}\)
\begin{equation*}
\begin{aligned} 15y^2\amp = \alert{1y(15y)}\\ \amp = 1y^2(15)\\ \amp = 1(15y^2)\\ \amp= 3y(5y)\\ \amp = 3(5y^2)\\ \amp = 3y^2(5)\\ \end{aligned}
\end{equation*}
In this case, the sum of the factors \(1y\) and \(15y\) equals the middle coefficient, \(16y\text{.}\) Therefore, we replace \(16xy\) with \(15xy+1xy\text{.}\)
\begin{equation*}
\begin{aligned} 5x^2+16xy+3y^2\amp = 5x^2+15xy+1xy+3y^2\\ \amp =5x(x+3y)+1y(x+3y)\\ \amp = (5x+1y)(x+3y)\\ \amp = (5x+y)(x+3y) \end{aligned}
\end{equation*}
Our factored form is \((5x+y)(x+3y)\text{.}\)
Example 4.42.
Factor \(18a^2b^2-ab-4\text{.}\)
Solution.
Here \(a=18,\, b=-1\text{,}\) and \(c=-4\text{.}\)
\begin{equation*}
\begin{aligned} ac=18(-4)=-72 \end{aligned}
\end{equation*}
Now we factor \(-72\text{,}\) and search for factors whose sum is \(-1\text{.}\)
\begin{equation*}
\begin{aligned} -72\amp = -1(72)\\ \amp = 1(-72)\\ \amp = -2(36)\\ \amp = 2(-36)\\ \amp = -4(18)\\ \amp = 4(-18)\\ \amp = -8(9)\\ \amp = \alert{8(-9)}\\ \amp = -24(3)\\ \amp = 24(-3)\\ \end{aligned}
\end{equation*}
In this case, the sum of the factors \(8\) and \(-9\) equals the middle coefficient, \(-1\text{.}\) Therefore, we replace \(-ab\) with \(8ab-9ab\text{.}\)
\begin{equation*}
\begin{aligned} 18a^2b^2-ab-4\amp = 18a^2b^2+(8ab-9ab)-4\\ \amp =18a^2b^2+(-9ab+8ab)-4\\ \amp =18a^2b^2-9ab+8ab-4\\ \amp =9ab(2ab-1)-4(2ab-1)\\ \amp = (9ab+4)(2ab-1)\\ \end{aligned}
\end{equation*}
Our factored form is \((9ab+4)(2ab-1)\text{.}\)
Example 4.43.
Factor \(12y^3-26y^2-10y\text{.}\)
Solution.
First, we will factor out the GCF.
\begin{equation*}
\begin{aligned} 12y^3-26y^2-10y=2y(6y^2-13y-5) \end{aligned}
\end{equation*}
After factoring out \(2y\text{,}\) the coefficients of the resulting trinomial are smaller and have fewer factors. We can factor the resulting trinomial using \(6=2(3)\) and \(5=(5)(1)\text{.}\) Notice that these factors can produce \(-13\) in two ways:
\begin{equation*}
\begin{aligned} 2(-5)+3(-1)=-10-3\amp =-13\\ 2(\alert{1})+3(\alert{-5})=2-15\amp =-13 \end{aligned}
\end{equation*}
Since the last term is \(-5\text{,}\) the correct combination requires the factors \(1\) and \(5\) to be opposite signs. Here we use \(2(1) = 2\) and \(3(-5) = -15\) because the sum is \(-13\) and the product of \((1)(-5) = -5\text{.}\)
\begin{equation*}
\begin{aligned} 12y^3-26y^2-10y\amp =2y(6y^2-13y-5)\\ \amp = 2y(2y \: \: )(3y\: \: )\\ \amp = 2y(2y-5)(3y+1) \end{aligned}
\end{equation*}
We check as follows.
\begin{equation*}
\begin{aligned} 2y(2y-5)(3y+1) \amp =2y(6y^2+2y-15y-5)\\ \amp = 2y(6y^2-13y-5)\\ \amp =12y^3-26y^2-10y \end{aligned}
\end{equation*}
The factor \(2y\) is part of the factored form of the original expression; be sure to include it in the answer.
Our factored form is \(2y(2y-5)(3y+1)\text{.}\)
Example 4.44.
Factor \(2x^2-9x-5\) using the \(ac\)-method.
Solution.
Since \(ac=(2)(-5)=-10\) and \(b=-9\text{,}\) we need two numbers that multiply to \(-10\) and sum to \(-9\text{.}\) Two such numbers are \(-10\) and \(1\text{,}\) so we can write \(-9x\) as \(-10x+1x\) and factor by grouping.
\begin{equation*}
\begin{aligned} 2x^2-9x-5 \amp= 2x^2-10x+1x-5 \\ \amp= 2x(x-5)+1(x-5) \\ \amp= (x-5)(2x+1) \end{aligned}
\end{equation*}
Example 4.45.
Factor \(6x^2+7x-10\) using the \(ac\)-method.
Solution.
Since \(ac=(6)(-10)=-60\) and \(b=7\text{,}\) we need two numbers that multiply to \(-60\) and sum to \(7\text{.}\) Two such numbers are \(12\) and \(-5\text{,}\) so we can write \(7x\) as \(7x=12x-5x\) and factor by grouping.
\begin{equation*}
\begin{aligned} 6x^2+7x-10 \amp= 6x^2+12x-5x-10 \\ \amp= 6x(x+2)-5(x+2) \\ \amp= (x+2)(6x-5) \end{aligned}
\end{equation*}
Checkpoint 4.46.
Factor \(9x^2+6x-8\) using the \(ac\)-method.
Exercises Exercises
1. Factoring Trinomials.
This is similar to Problem 2 from your Section 4.4 worksheet.
Factor.
\(x^2+15 x +56= \big(\) \(\big) \big(\) \(\big)\)
\(x^2-5 x - 24 = \big(\) \(\big) \big(\) \(\big)\)
\(x^2+18 x + 81 = \big(\) \(\big) \big(\) \(\big)\)
Hint.
These are all trinomials and require the ac-method.
2. Factoring Trinomials.
This is similar to Problem 2 from your Section 4.4 worksheet.
Factor.
\(5 x^2 + 5 x-60=\big(\) \(\big) \big(\) \(\big) \big(\) \(\big)\) (Note: The constant factor MUST be the first factor.)
\(x^2+15 xy + 56 y^2 = \big(\) \(\big) \big(\) \(\big)\)
\(m^2p^2-7 mp+10 = \big(\) \(\big) \big(\) \(\big)\)
Hint.
These are all trinomials and require the ac-method.
3. Factoring Trinomials.
This is similar to Problem 2 from your Section 4.4 worksheet.
Factor.
\(15 q^2-1 q-2= \big(\) \(\big) \big(\) \(\big)\)
\(12 x^2 + 41 x-11 = \big(\) \(\big) \big(\) \(\big)\)
\(m^4+16 m^2 +48 = \big(\) \(\big) \big(\) \(\big)\)
Hint.
These are all trinomials and require the ac-method.
