Learning Objectives

In this section students will:

5.3.1  – Factor by grouping.

5.3.2 – Factor a perfect square trinomial

5.3.3 – Factor a difference of squares.

5.3.1 – Factoring by Grouping

We have covered factoring trinomials with leading coefficient 1. In this section, we deal with the case when trinomials have leading coefficients other than 1.  In this case, factoring is slightly more complicated. 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 [latex]\,2{x}^{2}+5x+3\,[/latex] can be rewritten as [latex]\,\left(2x+3\right)\left(x+1\right)\,[/latex] using this process. We begin by rewriting the original expression as [latex]\,2{x}^{2}+2x+3x+3\,[/latex] and then factor each portion of the expression to obtain [latex]\,2x\left(x+1\right)+3\left(x+1\right).\,[/latex] We then pull out the GCF of [latex]\,\left(x+1\right)\,[/latex] to find the factored expression.

Factor by Grouping

To factor a trinomial in the form [latex]\,a{x}^{2}+bx+c\,[/latex] by grouping, we find two numbers with a product of [latex]\,ac\,[/latex] and a sum of [latex]\,b.\,[/latex] We use these numbers to divide the [latex]\,x\,[/latex] 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 [latex]\,a{x}^{2}+bx+c,[/latex] factor by grouping.

  1. List factors of [latex]\,ac.[/latex]
  2. Find [latex]\,p\,[/latex] and [latex]\,q,[/latex] a pair of factors of [latex]\,ac\,[/latex] with a sum of [latex]\,b.[/latex]
  3. Rewrite the original expression as [latex]\,a{x}^{2}+px+qx+c.[/latex]
  4. Pull out the GCF of [latex]\,a{x}^{2}+px.[/latex]
  5. Pull out the GCF of [latex]\,qx+c.[/latex]
  6. Factor out the GCF of the expression.

Example 1 – Factoring a Trinomial by Grouping

Factor [latex]\,5{x}^{2}+7x-6\,[/latex] by grouping.

We have a trinomial with [latex]\,a=5,b=7,[/latex] and [latex]\,c=-6.\,[/latex] First, determine [latex]\,ac=-30.\,[/latex] We need to find two numbers with a product of [latex]\,-30\,[/latex] and a sum of [latex]\,7.\,[/latex] In (Figure), we list factors until we find a pair with the desired sum.

 
Factors of [latex]\,-30[/latex] Sum of Factors
[latex]1,-30[/latex] [latex]-29[/latex]
[latex]-1,30[/latex] 29
[latex]2,-15[/latex] [latex]-13[/latex]
[latex]-2,15[/latex] 13
[latex]3,-10[/latex] [latex]-7[/latex]
[latex]-3,10[/latex] 7

So [latex]\,p=-3\,[/latex] and [latex]\,q=10.[/latex]

[latex]$$\begin{array}{cc}5{x}^{2}-3x+10x-6 \hfill & \phantom{\rule{2em}{0ex}}\text{Rewrite the original expression as }a{x}^{2}+px+qx+c.\hfill \\ x\left(5x-3\right)+2\left(5x-3\right)\hfill & \phantom{\rule{2em}{0ex}}\text{Factor out the GCF of each part}.\hfill \\ \left(5x-3\right)\left(x+2\right)\hfill & \phantom{\rule{2em}{0ex}}\text{Factor out the GCF}\text{​}\text{ of the expression}.\hfill \end{array}$$[/latex]

Analysis

We can check our work by multiplying. Use FOIL to confirm that [latex]\,\left(5x-3\right)\left(x+2\right)=5{x}^{2}+7x-6.[/latex]

Try It

Factor

a. [latex]\,2{x}^{2}+9x+9\,[/latex]

b. [latex]\,6{x}^{2}+x-1[/latex]

Show answer

a. [latex]\,\left(2x+3\right)\left(x+3\right)\,[/latex]

b. [latex]\,\left(3x-1\right)\left(2x+1\right)[/latex]

 

 5.3.2 – Factoring a Perfect Square Trinomial

A perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.

[latex]$$\begin{array}{ccc}\hfill {a}^{2}+2ab+{b}^{2}& =& {\left(a+b\right)}^{2}\hfill \\ & \text{and}& \\ \hfill {a}^{2}-2ab+{b}^{2}& =& {\left(a-b\right)}^{2}\hfill \end{array}$$[/latex]

We can use this equation to factor any perfect square trinomial.

Perfect Square Trinomials

A perfect square trinomial can be written as the square of a binomial:

[latex]$${a}^{2}+2ab+{b}^{2}={\left(a+b\right)}^{2}$$[/latex]

How To

Given a perfect square trinomial, factor it into the square of a binomial.

  1. Confirm that the first and last term are perfect squares.
  2. Confirm that the middle term is twice the product of [latex]\,ab.[/latex]
  3. Write the factored form as [latex]\,{\left(a+b\right)}^{2}.[/latex]

Example 2 – Factoring a Perfect Square Trinomial

Factor [latex]\,25{x}^{2}+20x+4.[/latex]

Notice that [latex]\,25{x}^{2}\,[/latex] and [latex]\,4\,[/latex] are perfect squares because [latex]\,25{x}^{2}={\left(5x\right)}^{2}\,[/latex] and [latex]\,4={2}^{2}.\,[/latex] Then check to see if the middle term is twice the product of [latex]\,5x\,[/latex] and [latex]\,2.\,[/latex] The middle term is, indeed, twice the product: [latex]\,2\left(5x\right)\left(2\right)=20x.\,[/latex] Therefore, the trinomial is a perfect square trinomial and can be written as [latex]\,{\left(5x+2\right)}^{2}.[/latex]

 

Try It

Factor [latex]\,49{x}^{2}-14x+1.[/latex]

Show answer

[latex]{\left(7x-1\right)}^{2}[/latex]

 

5.3.3 – Factoring a Difference of Squares

A difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied.

[latex]$${a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)$$[/latex]

We can use this equation to factor any differences of squares.

Differences of Squares

A difference of squares can be rewritten as two factors containing the same terms but opposite signs.

[latex]$${a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)$$[/latex]

How To

Given a difference of squares, factor it into binomials.

  1. Confirm that the first and last term are perfect squares.
  2. Write the factored form as [latex]\,\left(a+b\right)\left(a-b\right).[/latex]

Example 3 – Factoring a Difference of Squares

Factor [latex]\,9{x}^{2}-25.[/latex]

Notice that [latex]\,9{x}^{2}\,[/latex] and [latex]\,25\,[/latex] are perfect squares because [latex]\,9{x}^{2}={\left(3x\right)}^{2}\,[/latex] and [latex]\,25={5}^{2}.\,[/latex] The polynomial represents a difference of squares and can be rewritten as [latex]\,\left(3x+5\right)\left(3x-5\right).[/latex]

 

Try It

Factor [latex]\,81{y}^{2}-100.[/latex]

Show answer

[latex]\left(9y+10\right)\left(9y-10\right)[/latex]

 

Is there a formula to factor the sum of squares?

No. A sum of squares cannot be factored.

Access these online resources for additional instruction and practice with factoring polynomials.

Key Equations

difference of squares [latex]{a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)[/latex]
perfect square trinomial [latex]{a}^{2}+2ab+{b}^{2}={\left(a+b\right)}^{2}[/latex]
sum of cubes [latex]{a}^{3}+{b}^{3}=\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right)[/latex]
difference of cubes [latex]{a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)[/latex]
  • Trinomials can be factored using a process called factoring by grouping. See (Example 1).
  • Perfect square trinomials and the difference of squares are special products and can be factored using equations. See (Example 2) and (Example 3).

Glossary

factor by grouping
a method for factoring a trinomial in the form [latex]\,a{x}^{2}+bx+c\,[/latex] 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
greatest common factor
the largest polynomial that divides evenly into each polynomial

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