We've updated our
Privacy Policy effective December 15. Please read our updated Privacy Policy and tap

Study Guides > College Algebra CoRequisite Course

Factoring Basics

Learning Outcomes

  • Identify and factor the greatest common factor of a polynomial.
  • Factor a trinomial with leading coefficient 1.
  • Factor by grouping.
Recall that the greatest common factor (GCF) of two numbers is the largest number that divides evenly into both numbers. For example, 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}. Finding and factoring out a GCF from a polynomial is the first skill involved in factoring polynomials.

Factoring a GCF out of a polynomial

When factoring a polynomial expression, our first step is to check to see if each term contains a common factor. If so, we factor out the greatest amount we can from each term. To make it less challenging to find this GCF of the polynomial terms, first 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 a polynomial is the largest polynomial that divides evenly into each term of the polynomial.
To factor out a GCF from a polynomial, first identify the greatest common factor of the terms. You can then use the distributive property "backwards" to rewrite the polynomial in a factored form. Recall that the distributive property of multiplication over addition states that a product of a number and a sum is the same as the sum of the products.

Distributive Property Forward and Backward

Forward: We distribute aa over b+cb+c.

a(b+c)=ab+aca\left(b+c\right)=ab+ac.

Backward: We factor aa out of ab+acab+ac.

ab+ac=a(b+c)ab+ac=a\left(b+c\right).

We have seen that we can distribute a factor over a sum or difference. Now we see that we can "undo" the distributive property with factoring.

Example

Factor 25b3+10b225b^{3}+10b^{2}.

Answer: Find the GCF.

  25b3=55bbb  10b2=52bbGCF=5bb=5b2\begin{array}{l}\,\,25b^{3}=5\cdot5\cdot{b}\cdot{b}\cdot{b}\\\,\,10b^{2}=5\cdot2\cdot{b}\cdot{b}\\\text{GCF}=5\cdot{b}\cdot{b}=5b^{2}\end{array}

Rewrite each term with the GCF as one factor.

25b3=5b25b10b2=5b22\begin{array}{l}25b^{3} = 5b^{2}\cdot5b\\10b^{2}=5b^{2}\cdot2\end{array}

Rewrite the polynomial using the factored terms in place of the original terms.

5b2(5b)+5b2(2)5b^{2}\left(5b\right)+5b^{2}\left(2\right)

Factor out the 5b25b^{2}.

5b2(5b+2)5b^{2}\left(5b+2\right)

Answer

5b2(5b+2)5b^{2}\left(5b+2\right)

Analysis

The factored form of the polynomial 25b3+10b225b^{3}+10b^{2} is 5b2(5b+2)5b^{2}\left(5b+2\right). You can check this by doing the multiplication. 5b2(5b+2)=25b3+10b25b^{2}\left(5b+2\right)=25b^{3}+10b^{2}. Note that if you do not factor the greatest common factor at first, you can continue factoring, rather than start all over. For example:

25b3+10b2=5(5b3+2b2)           Factor out 5.                             =5b2(5b+2)               Factor out b2.\begin{array}{l}25b^{3}+10b^{2}=5\left(5b^{3}+2b^{2}\right)\,\,\,\,\,\,\,\,\,\,\,\text{Factor out }5.\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=5b^{2}\left(5b+2\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{Factor out }b^{2}.\end{array}

Notice that you arrive at the same simplified form whether you factor out the GCF immediately or if you pull out factors individually.

In the following video we see two more examples of how to find and factor the GCF from binomials. https://youtu.be/25_f_mVab_4

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 of the form xn{x}^{n} will always be the lowest exponent.) 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.

The GCF may not always be a monomial. Here is an example of a GCF that is a binomial.

Try It

Factor x(b2a)+6(b2a)x\left({b}^{2}-a\right)+6\left({b}^{2}-a\right) by pulling out the GCF.

Answer: (b2a)(x+6)\left({b}^{2}-a\right)\left(x+6\right)

[ohm_question]7888[/ohm_question]
Watch this video to see more examples of how to factor the GCF from a trinomial. https://youtu.be/3f1RFTIw2Ng

Factoring a Trinomial with Leading Coefficient 1

Although we should always begin by looking for a GCF, pulling out the GCF is not the only way that polynomial expressions can be factored. The polynomial x2+5x+6{x}^{2}+5x+6 has a GCF of 1, but it can be written as the product of the factors (x+2)\left(x+2\right) and (x+3)\left(x+3\right). Trinomials of the form x2+bx+c{x}^{2}+bx+c can be factored by finding two numbers with a product of cc and a sum of bb. The trinomial x2+10x+16{x}^{2}+10x+16, for example, can be factored using the numbers 22 and 88 because the product of these numbers is 1616 and their sum is 1010. The trinomial can be rewritten as the product of (x+2)\left(x+2\right) and (x+8)\left(x+8\right).

A General Note: Factoring a Trinomial with Leading Coefficient 1

A trinomial of the form x2+bx+c{x}^{2}+bx+c can be written in factored form as (x+p)(x+q)\left(x+p\right)\left(x+q\right) where pq=cpq=c and p+q=bp+q=b.

Q & A

Can every trinomial be factored as a product of binomials? No. Some polynomials cannot be factored. These polynomials are said to be prime.

How To: Given a trinomial in the form x2+bx+c{x}^{2}+bx+c, factor it

  1. List factors of cc.
  2. Find pp and qq, a pair of factors of cc with a sum of bb.
  3. Write the factored expression (x+p)(x+q)\left(x+p\right)\left(x+q\right).

Example: Factoring a Trinomial with Leading Coefficient 1

Factor x2+2x15{x}^{2}+2x - 15.

Answer: We have a trinomial with leading coefficient 1,b=21,b=2, and c=15c=-15. We need to find two numbers with a product of 15-15 and a sum of 22. In the table, we list factors until we find a pair with the desired sum.

Factors of 15-15 Sum of Factors
1,151,-15 14-14
1,15-1,15 1414
3,53,-5 2-2
3,5-3,5 22
Now that we have identified pp and qq as 3-3 and 55, write the factored form as (x3)(x+5)\left(x - 3\right)\left(x+5\right).

Analysis of the Solution

We can check our work by multiplying. Use FOIL to confirm that (x3)(x+5)=x2+2x15\left(x - 3\right)\left(x+5\right)={x}^{2}+2x - 15.

Q & A

Does the order of the factors matter? No. Multiplication is commutative, so the order of the factors does not matter.

Try It

Factor x27x+6{x}^{2}-7x+6.

Answer: (x6)(x1)\left(x - 6\right)\left(x - 1\right)

[ohm_question]7897[/ohm_question]

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: Factoring by Grouping

To factor a trinomial of 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+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+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

Factor the following.
  1. 2x2+9x+92{x}^{2}+9x+9
  2. 6x2+x16{x}^{2}+x - 1

Answer:

  1. (2x+3)(x+3)\left(2x+3\right)\left(x+3\right)
  2. (3x1)(2x+1)\left(3x - 1\right)\left(2x+1\right)

[ohm_question]7908[/ohm_question]
In the next video we see another example of how to factor a trinomial by grouping. [embed]https://youtu.be/agDaQ_cZnNc[/embed]

Licenses & Attributions