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Study Guides > College Algebra

Factoring by Grouping

Trinomials with leading coefficients other than 1 are slightly more complicated to factor. For these 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.

A General Note: Factor by Grouping

To factor a trinomial in the form ax2+bx+ca{x}^{2}+bx+c by grouping, we find two numbers with a product of acac and a sum of bb. We use these numbers to divide the xx term into the sum of two terms and factor each portion of the expression separately, then factor out the GCF of the entire expression.

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.

Example 3: Factoring a Trinomial by Grouping

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

Solution

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 29
2,152,-15 13-13
2,15-2,15 13
3,103,-10 7-7
3,10-3,10 7
So p=3p=-3 and q=10q=10.
5x23x+10x6Rewrite the original expression as ax2+px+qx+c.x(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}

Analysis of the Solution

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.

Try It 3

Factor the following.

a.2x2+9x+92{x}^{2}+9x+9 b. 6x2+x16{x}^{2}+x - 1

Solution

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