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<h4>Activity (20 minutes)</h4>
<p>This activity prompts students to learn an efficient method to multiply special polynomials. Students will learn to recognize multiplication expressions involving binomial squares and conjugate pairs. They will learn the patterns used to make the multiplication process more efficient than other methods used so far in this lesson.</p>
<h4> Launch</h4>
<p>Before starting the exercises, ask students to name any polynomial expression that can be identified as a binomial square.</p>
<p>Example of a binomial square: \((x + 3)^2\)</p>
<p>Ask them to name a conjugate pair.</p>
<p>Example of a conjugate pair: \((x - 5)(x + 5)\)</p>
<p>Ask students how to discern between the two different expressions. Then allow students to work through the exercises provided individually.</p>
<h4>Student Activity </h4>
<p>Multiply each expression. Identify each as a binomial square or a conjugate pair (difference of squares). Use the "^" symbol to enter an exponent.</p>
<ol class="os-raise-noindent">
<li> \((3x - 7)^2\) </li>
</ol>
<p><strong>Answer:</strong> \(9x^2 - 42x + 49\); binomial square</p>
<ol class="os-raise-noindent" start="2">
<li> \((6z + 4)(6z - 4)\) </li>
</ol>
<p><strong>Answer:</strong> \(36z^2 - 16\); conjugate pair</p>
<ol class="os-raise-noindent" start="3">
<li> \((4a + 2b)(4a + 2b)\) </li>
</ol>
<p><strong>Answer:</strong> \(16a^2 + 16ab + 4b^2\); binomial square</p>
<ol class="os-raise-noindent" start="4">
<li>\(
(3w - 9p)(3w + 9p)\) </li>
</ol>
<p><strong>Answer:</strong> \(9w^2 - 81p^2\); conjugate pair</p>
<ol class="os-raise-noindent" start="5">
<li> \((2y - 8n)(2y + 8n)\) </li>
</ol>
<p><strong>Answer:</strong> \(4y^2 - 64n^2\); conjugate pair</p>
<ol class="os-raise-noindent" start="6">
<li>\((5k + 2)(5k + 2)\) </li>
</ol>
<p><strong>Answer:</strong> \(25k^2 + 20k + 4\); binomial square</p>
<br>
<h4>Activity Synthesis</h4>
<p>Invite a student to name the equation used to identify the multiplication of a binomial square involving addition.</p>
<p>\((a + b)^2 = a^2 + 2ab +b^2\)</p>
<p>Ask another student how the equation involving subtraction is different.</p>
<p>\((a - b)^2 = a^2 - 2ab +b^2\)</p>
<p>Finally, ask a student to name the equation representing the multiplication of conjugate pairs.</p>
<p>\((a +b)(a - b) = a^2 - b^2\)</p>
<p>Ask students how mastering these equations makes multiplying these types of polynomials easier. Tell students that in future lessons, we will work on reversing this process using factoring.</p>
<br>
<h3>6.2.4: Self Check</h3>
<!--BEGIN SELF CHECK INTRO BEFORE Tables -->
<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>
<!--SELF CHECK QUESTION GOES BEFORE THE Table -->
<p><strong>QUESTION:</strong> </p>
<p>Multiply.</p>
<p>\((3f - 6g)^2\)</p>
<!--SELF CHECK table-->
<table class="os-raise-textheavytable">
<thead>
<tr>
<th scope="col">Answers</th>
<th scope="col">Feedback</th>
</tr>
</thead>
<tbody>
<tr>
<td><p><strong>Answers</strong></p></td>
<td><p><strong>Feedback</strong></p></td>
</tr>
<tr>
<td><p>\(9f^2-18fg+36g^2\)</p></td>
<td><p>Incorrect. Let’s try again a different way: You must multiply the middle term by 2. The correct form for the product of a binomial square is \((a - b)^2 = a^2-2ab+b^2\). The answer is \(9f^2-36fg+36g^2\).</p></td>
</tr>
<tr>
<td><p>\(9f^2-36fg+36g^2\)</p></td>
<td><p>That’s correct! Check yourself: The correct form for the product of a binomial square is \((a - b)^2 = a^2-2ab+b^2\).</p></td>
</tr>
<tr>
<td><p>\(9f^2-36fg-36g^2\)</p></td>
<td><p>Incorrect. Let’s try again a different way: Check the sign on the last term. The correct form for the product of a binomial square is \((a - b)^2 = a^2-2ab+b^2\). The answer is \(9f^2-36fg+36g^2\).</p></td>
</tr>
<tr>
<td><p>\(9f^2+36g^2\)</p></td>
<td><p>Incorrect. Let’s try again a different way: The middle term is missing. The correct form for the product of a binomial square is \((a - b)^2 = a^2-2ab+b^2\). The answer is \(9f^2-36fg+36g^2\).</p></td>
</tr>
</tbody>
</table>
<br>
<h3>Additional Resources</h3>
<p><em><strong>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.</strong></em></p>
<h4>Multiplying Binomial Squares</h4>
<p>Mathematicians like to look for patterns that will make their work easier. A good example of this is squaring binomials. While you can always get the product by writing the binomial twice and multiplying the two binomials, there is less work to do if you learn to use a pattern. Let’s start by looking at three examples and look for a pattern.</p>
<p>Look at these results. Do you see any patterns?</p>
<p><img height="189" src="https://k12.openstax.org/contents/raise/resources/3ed3cc403d17256f8e1249ea9c5eb5f5e9d1e75b" width="475"></p>
<p>What about the number of terms? In each example, we squared a binomial, and the result was a trinomial.</p>
<p>\((a + b)^2\) = ____ + ____ + ____</p>
<p>Now look at the <em>first term</em> in each result. Where did it come from?</p>
<p>The first term is the product of the first terms of each binomial. Since the binomials are identical, it is just the square of the first term!</p>
<p>\((a + b)^2 = a^2\) + ____ + ____</p>
<p><em>To get the first term of the product, square the first term.</em></p>
<p>Where did the <em>last term</em> come from? Look at the examples and find the pattern.</p>
<p>The last term is the product of the last terms, which is the square of the last term.</p>
<p>\((a + b)^2\) = ____ + ____ + \(b^2\)</p>
<p><em>To get the last term of the product, square the last term.</em></p>
<p>Finally, look at the <em>middle term</em>. Notice it came from adding the “outer” and the “inner” terms—which are the same! So the middle term is double the product of the two terms of the binomial.</p>
<p>\((a + b)^2\) = ____ \(+ \; 2ab \; +\) ____</p>
<p>\((a + b)^2\) = ____ \(- \; 2ab \; +\) ____</p>
<p><em>To get the middle term of the product, multiply the terms and double their product.</em></p>
<p>Putting it all together:</p>
<p><img height="253" src="https://k12.openstax.org/contents/raise/resources/3f2ec080f44dbc711dbc6af81aab127144c0431b" width="550"></p>
<h4>Try It: Multiplying Binomial Squares</h4>
<ol class="os-raise-noindent">
<li>Multiply \((x + 5)^2\).</li>
</ol>
<p><strong>Answer: </strong> Your answer may vary, but here is a sample.</p>
<p>\({\style{color:red}(}{\style{color:red}a}\;{\style{color:red}+}\;{\style{color:red}b}{\style{color:red})}^{\style{color:red}2}\)<br>
\((x+5)^2\) </p>
<p><strong>Step 1 - </strong> Square the first term.</p>
<p> \(\begin{array}{c}
{\style{color:red}a}
^
{\style{color:red}2}\;
{\style{color:red}+}\;
{\style{color:red}2}
{\style{color:red}a}
{\style{color:red}b}\;
{\style{color:red}+}\;
{\style{color:red}b}
^
{\style{color:red}2}
\\x^2+\_\_\_+\_\_\_\end{array}\)</p>
<p><strong>Step 2 - </strong> Square the last term.</p>
<p> \(\begin{array}{c}
{\style{color:red}a}
^
{\style{color:red}2}\;
{\style{color:red}+}\;
{\style{color:red}2}
{\style{color:red}a}
{\style{color:red}b}\;
{\style{color:red}+}\;
{\style{color:red}b}
^
{\style{color:red}2}
\\x^2+\_\_\_+5^2\end{array}\)</p>
<p><strong>Step 3 - </strong> Double the product of the two terms.</p>
<p> \(\begin{array}{c}
{\style{color:red}a}
^
{\style{color:red}2}\;
{\style{color:red}+}\;
{\style{color:red}2}\;
{\style{color:red}\cdot}\;
{\style{color:red}a}\;
{\style{color:red}\cdot}\;
{\style{color:red}b}\;
{\style{color:red}+}\;
{\style{color:red}b}
^
{\style{color:red}2}
\\x^2+2\cdot x\cdot5+5^2\end{array}\)</p>
<p><strong>Step 4 - </strong> Simplify.</p>
<p> \(x^2+10x+25\)</p>
<ol class="os-raise-noindent" start="2">
<li>Multiply \((2x - 3y)^2\).</li>
</ol>
<p><strong>Answer: </strong> Your answer may vary, but here is a sample.</p>
<p>\({\style{color:red}(}{\style{color:red}a}\;{\style{color:red}+}\;{\style{color:red}b}{\style{color:red})}^{\style{color:red}2}\)<br>
\((2x-3y)^2\) </p>
<p><strong>Step 1 - </strong> Use the pattern.<br>
\(\begin{array}{c}
{\style{color:red}a}
^
{\style{color:red}2}\;
{\style{color:red}-}\;
{\style{color:red}2}\;
{\style{color:red}\cdot}\;
{\style{color:red}a}\;
{\style{color:red}\cdot}\;
{\style{color:red}b}\;
{\style{color:red}+}\;
{\style{color:red}b}
^
{\style{color:red}2}
\\{(2x)}^2-2\cdot2x\cdot3y+{(3y)}^2\end{array}\)</p>
<p><strong>Step 2 - </strong> Simplify.<br>
\(4x^2-12xy+9y^2\)</p>
<br>
<h4>Conjugate Pairs</h4>
<p>We just saw a pattern for squaring binomials that we can use to make multiplying some binomials easier. Similarly, there is a pattern for another product of binomials.</p>
<p>A pair of binomials that each have the same first term and the same last term, but one is a sum and one is a difference, is called a <span class="os-raise-ib-tooltip" data-schema-version="1.0" data-store="glossary-tooltip">conjugate pair</span> and is of the form \((a-b)\), \((a+b)\).</p>
<div>
<p> <strong> CONJUGATE PAIR </strong> </p>
<hr>
<p>A conjugate pair is two binomials of the form</p>
<p>\((a-b),(a+b)\)</p>
<p>The pair of binomials each have the same first term and the same last term, but one binomial is a sum and the other is a difference.</p>
</div>
<br>
<p>There is a nice pattern for finding the product of conjugates. You could, of course, simply FOIL to get the product, but using the pattern makes your work easier. Let’s look for the pattern by using FOIL to multiply some conjugate pairs.</p>
<p><img height="124" src="https://k12.openstax.org/contents/raise/resources/b00d0cbbbb82c318af3fa32c414506b52cc17d55" width="475"></p>
<p>What do you observe about the products?</p>
<p>The product of the two binomials is also a binomial! Most of the products resulting from FOIL have been trinomials.</p>
<p>Each <em>first term</em> is the product of the first terms of the binomials, and since they are identical, it is the square of the first term.</p>
<p>\((a + b)(a - b) = a^2 -\) ____</p>
<p><em>To get the first term, square the first term.</em></p>
<p>The <em>last term</em> came from multiplying the last terms, and it is the square of the last term.</p>
<p>\((a + b)(a - b) = a^2 - b^2\)</p>
<p><em>To get the last term, square the last term.</em></p>
<p>Why is there no middle term? Notice the two middle terms you get from FOIL combine to 0 in every case, the result of one addition and one subtraction.</p>
<p>The product of conjugates is always of the form \(a^2 - b^2\). This is called a <span class="os-raise-ib-tooltip" data-schema-version="1.0" data-store="glossary-tooltip">difference of squares</span>.</p>
<p>This leads to the pattern:</p>
<div>
<p> <strong> PRODUCT OF CONJUGATES PATTERN </strong> </p>
<hr>
<p>If <em>a</em> and <em>b</em> are real numbers,</p>
<p><img alt class="img-fluid atto_image_button_text-bottom" height="273" role="presentation" src="https://k12.openstax.org/contents/raise/resources/f80744747c2668d03f93f976ce6253e0e17d86db" width="550"></p>
</div>
<br>
<p>We just developed special product patterns for Binomial Squares and for the Product of Conjugates. The products look similar, so it is important to recognize when it is appropriate to use each of these patterns and to notice how they differ. Look at the two patterns together and note their similarities and differences.</p>
<p><strong>COMPARING THE SPECIAL PRODUCT PATTERNS</strong></p>
<table class="os-raise-wideequaltable">
<thead>
<tr>
<th scope="col">Binomial Squares</th>
<th scope="col">Product of Conjugates</th>
</tr>
</thead>
<tbody>
<tr>
<td><p>\((a+b)^2=a^2+2ab+b^2\)</p></td>
<td><p>\((a-b)(a+b)=a^2-b^2\)</p></td>
</tr>
<tr>
<td><p>\((a-b)^2=a^2-2ab+b^2\)</p></td>
<td></td>
</tr>
<tr>
<td><p>Squaring a binomial.</p></td>
<td><p>Multiplying conjugates.</p></td>
</tr>
<tr>
<td><p>Product is a <strong>trinomial</strong>.</p></td>
<td><p>Product is a <strong>binomial</strong>.</p></td>
</tr>
<tr>
<td><p>Inner and outer terms with FOIL are <strong>the same</strong>.</p></td>
<td><p>Inner and outer terms with FOIL are <strong>opposites</strong>.</p></td>
</tr>
<tr>
<td><p>Middle term is <strong>double the product</strong> of the two terms.</p></td>
<td><p>There is <strong>no</strong> middle term.</p></td>
</tr>
</tbody>
</table>
<br>
<h4>Try It: Conjugate Pairs</h4>
<ol class="os-raise-noindent">
<li> Multiply \((2x + 5)(2x - 5)\).</li>
</ol>
<p><strong>Answer: </strong> Your answer may vary, but here is a sample.</p>
<p><strong>Step 1 - </strong> Are the binomials conjugates?</p>
<p> \((2x+5)(2x-5)\)<br>
<br>
It is the product of conjugates.</p>
<p> \(\begin{pmatrix}
{\style{color:red}a}\;
{\style{color:red}+}\;
{\style{color:red}b}
\\2x+5\end{pmatrix}\)
\(\begin{pmatrix}
{\style{color:red}a}\;
{\style{color:red}-}\;
{\style{color:red}b}
\\2x-5\end{pmatrix}\) </p>
<p><strong>Step 2 - </strong> Square the first term, \(2x\).</p>
<p> \(\begin{array}{c}
{\style{color:red}a}
^
{\style{color:red}2}\;
{\style{color:red}-}\;
{\style{color:red}b}
^
{\style{color:red}2}
\\{(2x)}^2-\_\_\_\end{array}\)</p>
<p><strong>Step 3 - </strong> Square the last term, 5.</p>
<p> \(\begin{array}{c}
{\style{color:red}a}
^
{\style{color:red}2}\;
{\style{color:red}-}\;
{\style{color:red}b}
^
{\style{color:red}2}
\\{(2x)}^2-5^2\end{array}\)</p>
<p><strong>Step 4 - </strong> Simplify. The product is a difference of squares.</p>
<p> \(\begin{array}{c}
{\style{color:red}a}
^
{\style{color:red}2}\;
{\style{color:red}-}\;
{\style{color:red}b}
^
{\style{color:red}2}
\\{4x^2}-25 \end{array}\) </p>
<ol class="os-raise-noindent" start="2">
<li> Multiply \((5m - 9n)(5m + 9n)\).</li>
</ol>
<p><strong>Answer: </strong> Your answer may vary, but here is a sample.</p>
<p><strong>Step 1 - </strong> This fits the pattern of the Product of Conjugates.</p>
<p> \(\begin{pmatrix}
{\style{color:red}a}\;
{\style{color:red}-}\:
{\style{color:red}b}
\\5m-9n\end{pmatrix}\begin{pmatrix}
{\style{color:red}a}\;
{\style{color:red}+}\;
{\style{color:red}b}
\\5m+9n\end{pmatrix}\\\)</p>
<p><strong>Step 2 - </strong> Use the pattern.</p>
<p> \(\begin{array}{c}
{\style{color:red}a}
^
{\style{color:red}2}\;
{\style{color:red}-}\;
{\style{color:red}b}
^
{\style{color:red}2}
\\{(5m)}^2-{(9n)}^2\end{array}\)</p>
<p><strong>Step 3 - </strong> Simplify.</p>
<p> \(25m^2-81n^2\)</p>