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Study Guides > College Algebra

Factoring Basics

Learning Objectives

  • Identify and factor the GCF of a polynomial
  • Factor a Trinomial with Leading Coefficient 1
  • Factor by Grouping
When we study fractions, we learn that the greatest common factor (GCF) of two numbers is the largest number that divides evenly into both numbers. For instance, 44 is the GCF of 1616 and 2020 because it is the largest number that divides evenly into both 1616 and 2020 The GCF of polynomials works the same way: 4x4x is the GCF of 16x16x and 20x220{x}^{2} because it is the largest polynomial that divides evenly into both 16x16x and 20x220{x}^{2}. When factoring a polynomial expression, our first step should be to check for a GCF. Look for the GCF of the coefficients, and then look for the GCF of the variables.

A General Note: Greatest Common Factor

The greatest common factor (GCF) of polynomials is the largest polynomial that divides evenly into the polynomials.

How To: Given a polynomial expression, factor out the greatest common factor.

  1. Identify the GCF of the coefficients.
  2. Identify the GCF of the variables.
  3. Combine to find the GCF of the expression.
  4. Determine what the GCF needs to be multiplied by to obtain each term in the expression.
  5. Write the factored expression as the product of the GCF and the sum of the terms we need to multiply by.

Example: Factoring the Greatest Common Factor

Factor 6x3y3+45x2y2+21xy6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy.

Answer: First, find the GCF of the expression. The GCF of 6,456,45, and 2121 is 33. The GCF of x3,x2{x}^{3},{x}^{2}, and xx is xx. (Note that the GCF of a set of expressions in the form xn{x}^{n} will always be the exponent of lowest degree.) And the GCF of y3,y2{y}^{3},{y}^{2}, and yy is yy. Combine these to find the GCF of the polynomial, 3xy3xy. Next, determine what the GCF needs to be multiplied by to obtain each term of the polynomial. We find that 3xy(2x2y2)=6x3y3,3xy(15xy)=45x2y23xy\left(2{x}^{2}{y}^{2}\right)=6{x}^{3}{y}^{3},3xy\left(15xy\right)=45{x}^{2}{y}^{2}, and 3xy(7)=21xy3xy\left(7\right)=21xy. Finally, write the factored expression as the product of the GCF and the sum of the terms we needed to multiply by.

(3xy)(2x2y2+15xy+7)\left(3xy\right)\left(2{x}^{2}{y}^{2}+15xy+7\right)

Analysis of the Solution

After factoring, we can check our work by multiplying. Use the distributive property to confirm that (3xy)(2x2y2+15xy+7)=6x3y3+45x2y2+21xy\left(3xy\right)\left(2{x}^{2}{y}^{2}+15xy+7\right)=6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy.

Try It

Factor x(b2−a)+6(b2−a)x\left({b}^{2}-a\right)+6\left({b}^{2}-a\right) by pulling out the GCF.

Answer: (b2−a)(x+6)\left({b}^{2}-a\right)\left(x+6\right)

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: Factoring a Trinomial by Grouping

Factor 5x2+7x−65{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 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.
5x2−3x+10x−6Rewrite the original expression as ax2+px+qx+c.x(5x−3)+2(5x−3)Factor out the GCF of each part.(5x−3)(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 (5x−3)(x+2)=5x2+7x−6\left(5x - 3\right)\left(x+2\right)=5{x}^{2}+7x - 6.

Try It

Factor the following.
  1. 2x2+9x+92{x}^{2}+9x+9
  2. 6x2+x−16{x}^{2}+x - 1

Answer:

  1. (2x+3)(x+3)\left(2x+3\right)\left(x+3\right)
  2. (3x−1)(2x+1)\left(3x - 1\right)\left(2x+1\right)

In the next video we show another example of how to factor a trinomial by grouping. https://youtu.be/agDaQ_cZnNc

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