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<h4>Activity (20 minutes)</h4>
<p>This activity introduces students to factoring a special product where a perfect square is subtracted from another
perfect square, \(a^2 − b^2\). The pattern that represents the factoring of this binomial is \(a^2 − b^2 =
(a − b)(a + b)\), where \((a − b)\) and \((a + b)\) are a conjugate pair.</p>
<h4>Launch</h4>
<p>Review the concept of conjugate pairs and introduce the difference of squares pattern as it is used to factor this
special case of a binomial.</p>
<p>Place students into pairs to work on the questions 1–7 in the activity. Allow students to work together to
factor the difference of squares.</p>
<p>Bring up question 7 to discuss as a class. Emphasize that the sum of two perfect squares does not match the pattern
given for the difference of squares and therefore cannot be factored.</p>
<div class="os-raise-extrasupport">
<div class="os-raise-extrasupport-header">
<p class="os-raise-extrasupport-title">Support for English Language Learners</p>
<p class="os-raise-extrasupport-name"><!--Extra Support Name-->MLR 3: Clarify, Critique, Correct</p>
</div>
<div class="os-raise-extrasupport-body">
<!--Support Content-->
<p>After discussing question 7 as a class, present an incorrect answer and explanation. For example, "\(49x^2+16y^2 = (7x+4y)(7x+4y)\) because when you find the square in a polynomial expression you take the square of each term." Ask students to identify the error, critique the reasoning, and write a correct explanation. Invite students to share their critiques and corrected explanations with the class. Listen for and amplify the language students use to explain how to find the difference of squares. This helps students evaluate, and improve upon, the written mathematical arguments of others. </p>
<p class="os-raise-text-italicize"><!--Support Italics-->Design Principle(s): Optimize output (for explanation); Maximize meta-awareness</p>
<p class="os-raise-extrasupport-title"><!--Extra Support Title-->Learn more about this routine</p>
<!-- Extra Support Content-->
<p><a href="https://www.youtube.com/watch?v=lozZJ21i3mQ;&rel=0" target="_blank">View the instructional video</a> and <a href="https://k12.openstax.org/contents/raise/resources/36e324584ae01e4eafc6674ac4ec02b4ae99002c" target="_blank">follow along with the materials</a> to assist you with learning this routine. </p>
<p class="os-raise-extrasupport-title"><!--Extra Support Title-->Provide support for students</p>
<!-- Extra Support Content-->
<p><a href="https://k12.openstax.org/contents/raise/resources/76d1756428038b737f48dc515486cddceb99e2f8" target="_blank">Distribute graphic organizers</a> to the students to assist them with participating in this routine. </p>
</div>
</div>
<br>
<h4>Activity</h4>
<p>Remember that \((a − b)\) and \((a + b)\) are a conjugate pair. When you multiply them together, you end up
with the difference of squares, \(a^2 − b^2\).</p>
<p>Let’s look at the pattern used to factor a difference of squares:</p>
<p>\(a^2 − b^2 = (a − b)(a + b)\)</p>
<p>Work with a partner to factor using the difference of squares. Use the "^" symbol to enter an exponent.</p>
<p>As you factor each binomial, take turns with your partner, alternating the completion of each step one at a time. If
you do not know the next step, ask your partner for help.</p>
<ol class="os-raise-noindent">
<li> \(81y^2 − 1\) </li>
</ol>
<p><strong>Answer:</strong> \((9y − 1)(9y + 1)\)</p>
<ol class="os-raise-noindent" start="2">
<li> \(196m^2 − 25n^2\) </li>
</ol>
<p><strong>Answer:</strong> \((14m − 5n)(14m + 5n)\)</p>
<ol class="os-raise-noindent" start="3">
<li> \(6p^2q^2 − 54p^2\) </li>
</ol>
<p><strong>Answer:</strong> \(6p^2(q − 3)(q + 3)\)</p>
<ol class="os-raise-noindent" start="4">
<li> \(7a^4c^2 − 7b^4c^2\) </li>
</ol>
<p><strong>Answer:</strong> \(7c^2(a − b)(a + b)(a^2 + b^2)\)</p>
<ol class="os-raise-noindent" start="5">
<li> \(a^2 + 6a + 9 − 9b^2\) </li>
</ol>
<p><strong>Answer:</strong> \((a + 3 − 3b)(a + 3 + 3b)\)</p>
<ol class="os-raise-noindent" start="6">
<li> \(x^2 + 6x + 9 − 4y^2\) </li>
</ol>
<p><strong>Answer:</strong> \((x + 3 − 2y)(x + 3 + 2y)\)</p>
<ol class="os-raise-noindent" start="7">
<li> Can a binomial with an addition operation be factored using the difference of squares? Explain. </li>
</ol>
<p><strong>Answer:</strong> No, the binomial that can be factored using a difference of squares will always have the
form \((a^2 − b^2)\).</p>
<h4>Activity Synthesis</h4>
<p>Review the following key points from the lesson:</p>
<ul>
<li> The difference of squares pattern is \(a^2 − b^2 = (a − b)(a + b)\), where the difference of squares
is factored into a conjugate pair. </li>
<li> After factoring a difference of squares into binomials, multiply the binomials together to check your answer. </li>
<li> The addition of two perfect squares in a binomial of the form \(a^2 + b^2\) is prime. </li>
</ul>
<h4>Video: Factoring the Difference of Squares</h4>
<p>Watch the following video to learn more about how to factor the difference of squares when there is also a greatest
common factor.</p>
<div class="os-raise-d-flex-nowrap os-raise-justify-content-center">
<div class="os-raise-video-container">
<video controls="true" crossorigin="anonymous">
<source src="https://k12.openstax.org/contents/raise/resources/3dd4ea7de318dc0911be9212995411f6c406a778">
<track default="true" kind="captions" label="On" src="https://k12.openstax.org/contents/raise/resources/086bdd4741914a59f9365c92f251f58e225f0211" srclang="en_us">
https://k12.openstax.org/contents/raise/resources/3dd4ea7de318dc0911be9212995411f6c406a778 </video>
</div>
</div>
<br>
<br>
<h3>6.6.3: Self Check</h3>
<p class="os-raise-text-bold"><em>After the activity, students will answer the following question to check their
understanding of the
concepts explored in the activity.</em></p>
<p class="os-raise-text-bold">QUESTION:</p>
<p>Factor using the difference of squares: \(121p^2 − 9q^2\).</p>
<table class="os-raise-textheavytable">
<thead>
<tr>
<th scope="col">Answers</th>
<th scope="col">Feedback</th>
</tr>
</thead>
<tbody>
<tr>
<td> \((11p − 3q)(11p + 3q)\) </td>
<td> That’s correct! Check yourself: Multiply the conjugate pair to see if it matches the original binomial. </td>
</tr>
<tr>
<td> \((11q − 3p)(11q + 3p)\) </td>
<td> Incorrect. Let’s try again a different way: The variables are transposed. This is not equivalent to the
original binomial. The correct answer is \((11p − 3q)(11p + 3q)\). </td>
</tr>
<tr>
<td> \((11p − 3q)^2\) </td>
<td> Incorrect. Let’s try again a different way: Factoring using the difference of squares will result in a
conjugate pair with opposite signs. The correct answer is \((11p − 3q)(11p + 3q)\). </td>
</tr>
<tr>
<td> \((11p + 3q)^2\) </td>
<td> Incorrect. Let’s try again a different way: Factoring using the difference of squares will result in a
conjugate pair with opposite signs. The correct answer is \((11p − 3q)(11p + 3q)\). </td>
</tr>
</tbody>
</table>
<br>
<h3>6.6.3: Additional Resources</h3>
<p class="os-raise-text-bold"><em>The following content is available to students who would like more support based on
their experience with
the self check. Students will not automatically have access to this content, so you may wish to share it with
those who could benefit from it.</em></p>
<h4>Factoring the Difference of Squares</h4>
<p>The other special product you saw in the previous chapter was the Product of Conjugates pattern. You used this to
multiply two binomials that were conjugates. Here’s an example:</p>
<p>\(\;\;
{\style{color:red}(}
{\style{color:red}a}\;
{\style{color:red}-}\;
{\style{color:red}b}
{\style{color:red})}
{\style{color:red}(}
{\style{color:red}a}\;
{\style{color:red}+}\;
{\style{color:red}b}
{\style{color:red})}
\\ (3x-4)(3x+4)\)</p>
<p>\(\;\;\;\;\;
{\style{color:red}(}
{\style{color:red}a}
{\style{color:red})}
^
{\style{color:red}2}\;
{\style{color:red}-}\;{\style{color:red}(} {\style{color:red}b}
{\style{color:red})}
^
{\style{color:red}2}
\\ \;\;\;\;(3x)^2-(4)^2\)</p>
<p>\(\;\;\;\;\;\;9x^2-16\) </p>
<p>A difference of squares pattern factors to a product of conjugates.</p>
<br>
<div class="os-raise-graybox"> <p class="os-raise-text-bold"> DIFFERENCE OF SQUARES PATTERN </p>
<hr>
<p>If \(a\) and \(b\) are real numbers,</p>
<p><img height="135" src="https://k12.openstax.org/contents/raise/resources/baf9f51d27792a26b9a9bf030e122a514ea0f3ee" width="498"></p>
</div>
<br>
<br>
<p>Remember, “difference” refers to subtraction. So, to use this pattern, you must make sure you have a
binomial in which two squares are being subtracted.</p>
<div class="os-raise-graybox">
<p class="os-raise-text-bold">How to factor the difference of squares: </p>
<hr>
<p><strong>Step 1 - </strong>Does the binomial fit the pattern?</p>
<p> Is this a difference?</p>
<p> Are the first and last terms perfect squares?<br>
\(\;\;\;a^2\;-\;b^2\\\;\_\_\_-\_\_\_\)</p>
<p><strong>Step 2 - </strong>Write them as squares. </p>
<p> \({(a)}^2-{(b)}^2\)</p>
<p><strong>Step 3 - </strong>Write the product of conjugates. </p>
<p> \((a-b)(a+b)\)</p>
<p><strong>Step 4 - </strong>Check by multiplying.</p>
</div>
<br>
<p class="os-raise-text-bold">Example 1</p>
<p>Factor: \(64y^2 − 1\).</p>
<p><strong>Step 1 -</strong> Does the binomial fit the pattern?</p>
<ul>
<li> Is this a difference?<br>
Yes.</li>
<li> Are the first and last terms perfect squares?<br>
Yes.</li>
</ul>
<p><strong>Step 2 -
</strong>Write them as the difference of squares.</p>
<p> \(\;\;\;\;\;
{\style{color:red}a}
^
{\style{color:red}2}
{\style{color:red}-}
\;
{\style{color:red}b}
^
{\style{color:red}2}
\\ (8y)^2-1^2\)</p>
<p><strong>Step 3 -</strong> Write the product of conjugates.</p>
<p> \(\;
{\style{color:red}(}
{\style{color:red}a}\;
{\style{color:red}-}\;
{\style{color:red}b}
{\style{color:red})}\;
{\style{color:red}(}
{\style{color:red}a}\;
{\style{color:red}+}\;
{\style{color:red}b}
{\style{color:red})}\\ (8y-1)(8y+1)\)</p>
<p><strong>Step 4 -</strong> Check by multiplying.</p>
<p> \((8y − 1)(8y + 1)\)<br>
\(64y^2 − 1\; \checkmark\)</p>
<br>
<p>It is important to remember that sums of squares do not factor into a product of binomials. There are no binomial
factors that multiply together to get a sum of squares. After removing any GCF, the expression \(a^2 + b^2\) is prime! </p>
<p>The next example shows variables in both terms.</p>
<p class="os-raise-text-bold">Example 2</p>
<p>Factor: \(144x^2 − 49y^2\).</p>
<p><strong>Step 1 -
</strong>Is this a difference of squares?</p>
<p> \(144x^2 − 49y^2\)</p>
<p> Yes.</p>
<p> \((12x)^2 − (7y)^2\)</p>
<p><strong>Step 2 -</strong> Factor as the product of conjugates.</p>
<p> \((12x − 7y)(12x + 7y)\)</p>
<p><strong>Step 3 -</strong> Check by multiplying.</p>
<p> \((12x − 7y)(12x + 7y)\)<br>
\(44x^2 − 49y^2\; \checkmark\)</p>
<p>As always, you should look for a common factor first whenever you have an expression to factor. Sometimes a common
factor may “disguise” the difference of squares, and you won’t recognize the perfect squares until
you factor out the GCF.</p>
<p>Also, to completely factor the binomial in the next example, we’ll factor a difference of squares twice!</p>
<p class="os-raise-text-bold">Example 3</p>
<p>Factor: \(48x^4y^2 − 243y^2\).</p>
<p><strong>Step 1 -</strong> Is there a GCF?</p>
<p> \(48x^4y^2 − 243y^2\)<br>
Yes, \(3y^2\)—factor it out!<br>
\(3y^2(16x^4-81)\)</p>
<p><strong>Step 2 -</strong> Is the binomial a difference of squares? Yes.</p>
<p> \(3y^2[(4x^2)^2-(9)^2]\)</p>
<p><strong>Step 3 -</strong> Factor as a product of conjugates.</p>
<p> \(3y^2(4x^2-9)(4x^2+9)\)</p>
<p><strong>Step 4 -</strong> Notice the first binomial is also a difference of squares!</p>
<p> \(3y^2[(2x)^2-(3)^2](4x^2+9)\)</p>
<p><strong>Step 5 -</strong> Factor it as the product of conjugates.</p>
<p> \(3y^2(2x-3)(2x+3)(4x^2+9)\)</p>
<p> The last factor, the sum of squares, cannot be factored.</p>
<p><strong>Step 6 -</strong> Check by multiplying.</p>
<p> \(3y^2(2x-3)(2x+3)(4x^2+9)\)<br>
\(3y^2(4x^2-9)(4x^2+9)\)<br>
\(3y^2(16x^4-81)\)<br>
\(48x^4y^2 − 243y^2\; \checkmark\)</p>
<p class="os-raise-text-bold">Example 4</p>
<p>Factor: \(x^2 − 6x + 9 − y^2\).</p>
<p><strong>Step 1 -</strong> Factor by grouping the first three terms.</p>
<p><img height="46" src="https://k12.openstax.org/contents/raise/resources/8741ef515edd3b78f30f5c88bc4ea5e87489d641" width="213"> </p>
<p><strong>Step 2 -</strong> Use the perfect square trinomial pattern.</p>
<p> \((x − 3)^2 − y^2\).</p>
<p><strong>Step 3 -
</strong>Is there a difference of squares?</p>
<p> Yes—write them as squares.</p>
<p> \(\;\;\;\;\;
{\style{color:red}a}
^
{\style{color:red}2}
\;\;\; {\style{color:red}-}
\;
{\style{color:red}b}
^
{\style{color:red}2} \\ (x-3)^2
-y^2\) </p>
<p><strong>Step 4 -
</strong>Factor as the product of conjugates.</p>
<p> \(\;\;\;\;
{\style{color:red}(}
{\style{color:red}a}\;\;\;\; {\style{color:red}-}\;
{\style{color:red}b}
{\style{color:red})}\;\;\;
{\style{color:red}(}
{\style{color:red}a}\;\;\;\;\;
{\style{color:red}+}\;
{\style{color:red}b}
{\style{color:red})} \\ ((x-3)-y)((x-3)+y)\) <br>
\((x − 3 − y)(x − 3 + y)\) </p>
<p>You may want to rewrite the solution as \((x − y − 3)(x + y − 3)\).</p>
<h4>Try It: Factoring the Difference of Squares</h4>
<p>Factor using the difference of squares.</p>
<ol class="os-raise-noindent">
<li> \(121m^2 − 1\) </li>
</ol>
<ol class="os-raise-noindent" start="2">
<li> \(2x^4y^2 − 32y^2\) </li>
</ol>
<ol class="os-raise-noindent" start="3">
<li> \(x^2 − 10x + 25 − y^2\) </li>
</ol>
<p>Write down your answers. Then select the <strong>solution</strong> button to compare your work.</p>
<h5>Solution</h5>
<p>Here is how to factor trinomials using the difference of squares.</p>
<ol class="os-raise-noindent">
<li> \((11m − 1)(11m + 1)\) </li>
</ol>
<ol class="os-raise-noindent" start="2">
<li> \(2y^2(x − 2)(x + 2)(x^2 + 4)\) </li>
</ol>
<ol class="os-raise-noindent" start="3">
<li> \((x − 5 − y)(x − 5 + y)\) </li>
</ol>