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Study Guides > ALGEBRA / TRIG I

Factoring by Grouping

Learning Outcome

  • Apply an algorithm to rewrite a trinomial as a four term polynomial and factor
  • Use factoring by grouping to factor a trinomial
  • Factor trinomials of the form ax2+bx+ca{x}^{2}+bx+c
In the last section, we showed you how to factor polynomials with four terms by grouping.  Trinomials of the form ax2+bx+ca{x}^{2}+bx+c are slightly more complicated to factor. For trinomials, we can factor by grouping by dividing the x term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial 2x2+5x+32{x}^{2}+5x+3 can be rewritten as (2x+3)(x+1)\left(2x+3\right)\left(x+1\right) using this process. We begin by rewriting the original expression as 2x2+2x+3x+32{x}^{2}+2x+3x+3 and then factor each portion of the expression to obtain 2x(x+1)+3(x+1)2x\left(x+1\right)+3\left(x+1\right). We then pull out the GCF of (x+1)\left(x+1\right) to find the factored expression. Below is a summary of the steps we will use followed by an example demonstrating how to use the step.

Factoring Trinomials in the form ax2+bx+cax^{2}+bx+c

To factor a trinomial in the form ax2+bx+cax^{2}+bx+c, find two integers, r and s, whose sum is b and whose product is ac.

rs=acr+s=b\begin{array}{l}r\cdot{s}=a\cdot{c}\\r+s=b\end{array}

Rewrite the trinomial as ax2+rx+sx+cax^{2}+rx+sx+c and then use grouping and the distributive property to factor the polynomial.
The first step in this process is to figure out what two numbers to use to re-write the x term as the sum of two new terms. Making a table to keep track of your work is helpful. We are looking for two numbers with a product of 66 and a sum of 55
Factors of 23=62\cdot3=6 Sum of Factors
1,61,6 77
1,6-1,-6 7-7
2,32,3 55
2,3-2,-3 5-5
  The pair p=2 and q=3p=2 \text{ and }q=3 will give the correct x term, so we will rewrite it using the new factors:

2x2+5x+3=2x2+2x+3x+32{x}^{2}+5x+3=2x^2+2x+3x+3

Now we can group the polynomial into two binomials.

2x2+2x+3x+3=(2x2+2x)+(3x+3)2x^2+2x+3x+3=(2x^2+2x)+(3x+3)

Identify the GCF of each binomial.

2x2x is the GCF of (2x2+2x)(2x^2+2x) and 33 is the GCF of (3x+3)(3x+3). Use this to rewrite the polynomial:

(2x2+2x)+(3x+3)=2x(x+1)+3(x+1)(2x^2+2x)+(3x+3)=2x(x+1)+3(x+1)

Note how we leave the signs in the binomials and the addition that joins them. Be careful with signs when you factor out the GCF. The GCF of our new polynomial is (x+1)(x+1). We factor this out as well:

2x(x+1)+3(x+1)=(x+1)(2x+3)2x(x+1)+3(x+1)=(x+1)(2x+3).

Sometimes it helps visually to write the polynomial as (x+1)2x+(x+1)3(x+1)2x+(x+1)3 before you factor out the GCF. This is purely a matter of preference. Multiplication is commutative, so order does not matter.

Now let’s see how this strategy works for factoring 6z2+11z+46z^{2}+11z+4. In this trinomial, a=6a=6, b=11b=11, and c=4c=4. According to the strategy, you need to find two factors, r and s, whose sum is b=11b=11 and whose product is ac=64=24a\cdot{c}=6\cdot4=24. You can make a chart to organize the possible factor combinations. (Notice that this chart only has positive numbers. Since ac is positive and b is positive, you can be certain that the two factors you're looking for are also positive numbers.)
Factors whose product is  2424 Sum of the factors
124=241\cdot24=24 1+24=251+24=25
212=242\cdot12=24 2+12=142+12=14
38=243\cdot8=24 3+8=113+8=11
46=244\cdot6=24 4+6=104+6=10
There is only one combination where the product is 2424 and the sum is 1111, and that is when r=3r=3, and s=8s=8. Let’s use these values to factor the original trinomial.

Example

Factor 6z2+11z+46z^{2}+11z+4.

Answer: Rewrite the middle term, 11z11z, as 3z+8z3z + 8z (from the chart above.)

6z2+3z+8z+46z^{2}+3z+8z+4

Group pairs. Use grouping to consider the terms in pairs.

(6z2+3z)+(8z+4)\left(6z^{2}+3z\right)+\left(8z+4\right)

Factor 33z out of the first group and 44 out of the second group.

3z(2z+1)+4(2z+1)3z\left(2z+1\right)+4\left(2z+1\right)

Factor out (2z+1)\left(2z+1\right).

(2z+1)(3z+4)\left(2z+1\right)\left(3z+4\right)

Answer

(2z+1)(3z+4)\left(2z+1\right)\left(3z+4\right)

Now, let's look at an example where c is negative.

Example

Factor 5x2+7x65{x}^{2}+7x - 6 by grouping.

Answer: We have a trinomial with a=5,b=7a=5,b=7, and c=6c=-6. First, determine ac=30ac=-30. We need to find two numbers with a product of 30-30 and a sum of 77. In the table, we list factors until we find a pair with the desired sum.

Factors of 30-30 Sum of Factors
1,301,-30 29-29
1,30-1,30 2929
2,152,-15 13-13
2,15-2,15 1313
3,103,-10 7-7
3,10-3,10 77
So p=3p=-3 and q=10q=10. 5x23x+10x6Rewrite the original expression as ax2+px+qx+cx(5x3)+2(5x3)Factor out the GCF of each part(5x3)(x+2)Factor out the GCF  of the expression\begin{array}{cc}5{x}^{2}-3x+10x - 6 \hfill & \text{Rewrite the original expression as }a{x}^{2}+px+qx+c\hfill \\ x\left(5x - 3\right)+2\left(5x - 3\right)\hfill & \text{Factor out the GCF of each part}\hfill \\ \left(5x - 3\right)\left(x+2\right)\hfill & \text{Factor out the GCF}\text{ }\text{ of the expression}\hfill \end{array} We can check our work by multiplying. Use FOIL to confirm that (5x3)(x+2)=5x2+7x6\left(5x - 3\right)\left(x+2\right)=5{x}^{2}+7x - 6.

We can summarize our process in the following way:

How To: Given a trinomial in the form ax2+bx+ca{x}^{2}+bx+c, factor by grouping

  1. List factors of acac.
  2. Find pp and qq, a pair of factors of acac with a sum of bb.
  3. Rewrite the original expression as ax2+px+qx+ca{x}^{2}+px+qx+c.
  4. Pull out the GCF of ax2+pxa{x}^{2}+px.
  5. Pull out the GCF of qx+cqx+c.
  6. Factor out the GCF of the expression.

Try It

[ohm_question]115262[/ohm_question]
In the following video, we present another example of factoring a trinomial using grouping.  In this example, the middle term, b, is negative. Note how having a negative middle term and a positive c term influence the options for r and s when factoring.
https://youtu.be/agDaQ_cZnNc Using this strategy, factoring trinomials becomes quick and kind of fun once you get the idea.  Give the next examples a try on your own before you look at the solution.

We will show two more examples so you can become acquainted with the variety of possible outcomes for factoring this type of trinomial.

Example

Factor 2x2+9x+92{x}^{2}+9x+9.

Answer: Find two numbers p, q such that pq=18p\cdot{q}=18 and p+q=9p + q = 9. 99 and 1818 are both positive, so we will only consider positive factors.

Factors of 29=182\cdot9=18 Sum of Factors
1,181, 18 1919
3,63,6 99
We can stop because we have found our factors. Rewrite the original expression and group.

2x2+3x+6x+9=(2x2+3x)+(6x+9)2x^2+3x+6x+9=(2x^2+3x)+(6x+9)

Factor out the GCF of each binomial and write as a product of two binomials:

(2x2+3x)+(6x+9)=x(2x+3)+3(2x+3)=(x+3)(2x+3)(2x^2+3x)+(6x+9)=x(2x+3)+3(2x+3)=(x+3)(2x+3)

2x2+9x+9=(x+3)(2x+3)2{x}^{2}+9x+9=(x+3)(2x+3)

Here is an example for you to try where the c is negative.

Example

Factor 6x2+x16{x}^{2}+x - 1.

Answer:  

Factors of 61=66\cdot-1=-6 Sum of Factors
1,6-1,6 55
1,61,-6 5-5
2,3-2,3 11
We can stop because we have found our factors. Rewrite the original expression and group.

6x2+x1=6x22x+3x16{x}^{2}+x - 1=6x^2-2x+3x-1

Factor out the GCF of each binomial and write as a product of two binomials:

(6x22x)+(3x1)=2x(3x1)+1(3x1)=(2x+1)(3x1)(6x^2-2x)+(3x-1)=2x(3x-1)+1(3x-1)=(2x+1)(3x-1)

6x2+x1=(2x+1)(3x1)6{x}^{2}+x - 1=(2x+1)(3x-1)

Prime Trinomials

Before going any further, it is worth mentioning that not all trinomials can be factored using integer pairs. Take the trinomial 2z2+35z+72z^{2}+35z+7, for instance. Can you think of two integers whose sum is b=35b=35 and whose product is ac=27=14a\cdot{c}=2\cdot7=14? There are none! This type of trinomial, which cannot be factored using integers, is called a prime trinomial.  For our last example, you will see that sometimes, you will encounter polynomials that, despite your best efforts, cannot be factored into the product of two binomials.

Example

Factor 7x216x57x^{2}-16x–5.

Answer: Find p,qp, q such that pq=35 and p+q=16p\cdot{q}=-35\text{ and }p+q=-16

Factors of 75=357\cdot{-5}=-35 Sum of Factors
1,35-1, 35 3434
1,351, -35 34-34
5,7-5, 7 22
7,5-7,5 2-2
The trinomial cannot be factored. None of the factors add up to 16-16

Factoring out a GCF

Now that you are familiar with the grouping method for factoring, we will now start to look at some strategies that are useful for making factoring easier.  Your first step when factoring should always be to look for common factors for the three terms.  Consider the examples below.
Trinomial Factor out Common Factor Factored
2x2+10x+122x^{2}+10x+12 2(x2+5x+6)2(x^{2}+5x+6) 2(x+2)(x+3)2\left(x+2\right)\left(x+3\right)
5a215a10−5a^{2}−15a−10 5(a2+3a+2)−5(a^{2}+3a+2) 5(a+2)(a+1)−5\left(a+2\right)\left(a+1\right)
c38c2+15cc^{3}–8c^{2}+15c c(c28c+15)c\left(c^{2}–8c+15\right) c(c5)(c3)c\left(c–5\right)\left(c–3\right)
y49y310y2y^{4}–9y^{3}–10y^{2} y2(y29y10)y^{2}\left(y^{2}–9y–10\right) y2(y10)(y+1)y^{2}\left(y–10\right)\left(y+1\right)
Notice that once you have identified and pulled out the common factor, you can factor the remaining trinomial as usual. This process is shown below.

Example

Factor 3x33x290x3x^{3}–3x^{2}–90x.

Answer: Since 3 is a common factor for the three terms, factor out the 3.

3(x3x230x)3\left(x^{3}–x^{2}–30x\right)

x is also a common factor, so factor out x.

3x(x2x30)3x\left(x^{2}–x–30\right)

Now you can factor the trinomial x2x30x^{2}–x–30. To find r and s, identify two numbers whose product is 30−30 and whose sum is 1−1. The pair of factors is 6−6 and 55. So replace x–x with 6x+5x−6x+5x.

3x(x26x+5x30)3x\left(x^{2}–6x+5x–30\right)

Use grouping to consider the terms in pairs.

3x[(x26x)+(5x30)]3x\left[\left(x^{2}–6x\right)+\left(5x–30\right)\right]

Factor x out of the first group and factor 55 out of the second group.

3x[(x(x6))+5(x6)]3x\left[\left(x\left(x–6\right)\right)+5\left(x–6\right)\right]

Then factor out x6x–6.

3x(x6)(x+5)3x\left(x–6\right)\left(x+5\right)

Answer

3x(x6)(x+5)3x\left(x–6\right)\left(x+5\right)

The general form of trinomials with a leading coefficient of a is ax2+bx+cax^{2}+bx+c. Sometimes the factor of a can be factored as you saw above; this happens when a can be factored out of all three terms. The remaining trinomial that still needs factoring will then be simpler, with the leading term only being an x2x^{2} term, instead of an ax2ax^{2} term.

Factoring out a Negative

In some situations, a is negative, as in 4h2+11h+3−4h^{2}+11h+3. It often makes sense to factor out 1−1 as the first step in factoring, as doing so will change the sign of ax2ax^{2} from negative to positive, making the remaining trinomial easier to factor.

Example

Factor 4h2+11h+3−4h^{2}+11h+3

Answer: Factor 1−1 out of the trinomial. Notice that the signs of all three terms have changed.

1(4h211h3)−1\left(4h^{2}–11h–3\right)

To factor the trinomial, you need to figure out how to rewrite 11h−11h. The product of rs=43=12rs=4\cdot−3=−12, and the sum of rs=11rs=−11.
rs=12r\cdot{s}=−12 r+s=11r+s=−11
121=12−12\cdot1=−12 12+1=11−12+1=−11
62=12−6\cdot2=−12 6+2=4−6+2=−4
43=12−4\cdot3=−12 4+3=1−4+3=−1
Rewrite the middle term 11h−11h as 12h+1h−12h+1h.

1(4h212h+1h3)−1\left(4h^{2}–12h+1h–3\right)

Group terms.

1[(4h212h)+(1h3)]−1\left[\left(4h^{2}–12h\right)+\left(1h–3\right)\right]

Factor out 44h from the first pair. The second group cannot be factored further, but you can write it as +1(h3)+1\left(h–3\right) since +1(h3)=(h3)+1\left(h–3\right)=\left(h–3\right). This helps with factoring in the next step.

1[4h(h3)+1(h3)]−1\left[4h\left(h–3\right)+1\left(h–3\right)\right]

Factor out a common factor of (h3)\left(h–3\right). Notice you are left with (h3)(4h+1)\left(h–3\right)\left(4h+1\right); the +1+1 comes from the term +1(h3)+1\left(h–3\right) in the previous step.

1[(h3)(4h+1)]−1\left[\left(h–3\right)\left(4h+1\right)\right]

Answer

1(h3)(4h+1)−1\left(h–3\right)\left(4h+1\right)

Note that the answer above can also be written as (h+3)(4h+1)\left(−h+3\right)\left(4h+1\right) or (h3)(4h1)\left(h–3\right)\left(−4h–1\right) if you multiply 1−1 times one of the other factors. In the following video example, we will factor a trinomial whose leading term is negative.  You will see how, by factoring out the negative sign, factoring the trinomial becomes easier. https://youtu.be/zDAMjdBfkDs

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