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<h4>Activity (15 minutes)</h4>
<p>Most of the factored expressions students saw were of the form \((x + m)(x + n)\) or \(x(x + m)\). In this activity,
students work with expressions of the form \((px + m)(qx + n)\). They expand expressions such as \((2x + 1)(x + 10)\)
into standard form and look for structure that would allow them to go in reverse, that is, to transform expressions of
the form \(ax^2+ bx + c\)
</p>
<p>Going from factored form to standard form is fairly straightforward given students' experience with the distributive
property. Going in reverse, however, is a bit more challenging when the coefficient of \(x^2\) is not 1. With some
guessing and checking, students should be able to find the factored form of the expressions in the fourth and fifth
questions, but they should also notice that this process is not straightforward.</p>
<h4>Launch</h4>
<p>Remind students that they have seen quadratic expressions such as \(16t^2+ 800t + 400\) and \(5x^2+ 21x − 20\),
where the coefficient of the squared term is not 1. Solicit some ideas from students on how to write the factored form
for expressions such as these.</p>
<p>Arrange students in groups of two to three and ask them to split up the work for completing the first three
questions, with each group member rewriting one expression into standard form.</p>
<p>Invite students to share the expanded expressions in standard form. Record the expressions in the right column, and
ask students to make observations about them.</p>
<p>If not mentioned by students, point out that each pair of factors start with \(3x\) and \(x\), which multiply to make
\(3x^2\). Each pair of factors also has constant terms that multiply to make 4. The resulting expressions in standard
form are all different, however, because using different pairs of factors of 4 and arranging them in different orders
produce different expanded expressions.</p>
<p>Ask students to keep these observations in mind as they complete questions 4 and 5.</p>
<p>The student extension and the self check quiz contain examples in standard form where the trinomials have negative
\(c\)-values. Ask students how having a negative \(c\)-value would affect the way they approach factoring. (The
constants in the binomials must have opposite signs.)</p>
<br>
<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">MLR 8 Discussion Supports: Speaking, Representing</p>
</div>
<div class="os-raise-extrasupport-body">
<p>Give students time to make sure that everyone in the group can explain or justify each step or part of the problem. Invite groups to rehearse what they will say when they share with the whole class. Rehearsing provides students with additional opportunities to speak and clarify their thinking, and will improve the quality of explanations shared during the whole-class discussion. Then make sure to vary who is selected to represent the work of the group so that students get accustomed to preparing each other to fill that role.</p>
<p class="os-raise-text-italicize">Design Principle(s): Support sense-making; Cultivate conversation</p>
</div>
</div>
<br>
<div class="os-raise-extrasupport">
<div class="os-raise-extrasupport-header">
<p class="os-raise-extrasupport-title">Support for Students with Disabilities</p>
<p class="os-raise-extrasupport-name">Representation: Internalize Comprehension</p>
</div>
<div class="os-raise-extrasupport-body">
<p>Activate or supply background knowledge. Demonstrate how students can continue to use diagrams to rewrite expressions in which the coefficient of the squared term is not 1. Invite students to begin by generating a list of factors and to test them using the diagram. Encourage students to persist with this method, reiterating the fact that they are not necessarily expected to immediately recognize which factors will work without testing them. Allow students to use calculators to ensure inclusive participation.</p>
<p class="os-raise-text-italicize">Supports accessibility for: Visual-spatial processing; Organization</p>
</div>
</div>
<br>
<h4>Student Activity</h4>
<p>Each row in the given tables contains a pair of equivalent expressions. Complete the tables by converting the given
expression into standard form. If you get stuck, try drawing a diagram.</p>
<p>[Remember to use the ^ key to enter exponents. You can find this symbol by pushing shift and the number 6 on your
keyboard at the same time.]</p>
<ol class="os-raise-noindent">
<li></li>
</ol>
<table class="os-raise-midsizetable">
<thead>
<tr>
<th scope="col">
Factored form
</th>
<th scope="col">
Standard form
</th>
</tr>
</thead>
<tbody>
<tr>
<td>
<p>\((3x + 1)(x + 4)\)</p>
</td>
<td> </td>
</tr>
</tbody>
</table>
<br>
<p><strong>Answer:</strong> \(3x^2+13x +4\)</p>
<ol class="os-raise-noindent" start="2">
<li></li>
</ol>
<table class="os-raise-midsizetable">
<thead>
<tr>
<th scope="col">
Factored form
</th>
<th scope="col">
Standard form
</th>
</tr>
</thead>
<tbody>
<tr>
<td>
<p>\((3x + 2)(x + 2)\)</p>
</td>
<td> </td>
</tr>
</tbody>
</table>
<br>
<p><strong>Answer:</strong> \(3x^2 + 8x + 4\)</p>
<ol class="os-raise-noindent" start="3">
<li></li>
</ol>
<table class="os-raise-midsizetable">
<thead>
<tr>
<th scope="col">
Factored form
</th>
<th scope="col">
Standard form
</th>
</tr>
</thead>
<tbody>
<tr>
<td>
<p>\((3x + 4)(x + 1)\)</p>
</td>
<td> </td>
</tr>
</tbody>
</table>
<br>
<p><strong>Answer:</strong> \(3x^2 + 7x + 4\)</p>
<ol class="os-raise-noindent" start="4">
<li>To find the <span class="os-raise-ib-tooltip" data-schema-version="1.0" data-store="glossary-tooltip">factored
form</span> of a <span class="os-raise-ib-tooltip" data-schema-version="1.0"
data-store="glossary-tooltip">quadratic expression</span> with a leading <span class="os-raise-ib-tooltip"
data-schema-version="1.0" data-store="glossary-tooltip">coefficient</span> other than 1, answer parts a–d.
</li>
</ol>
<table class="os-raise-midsizetable">
<thead>
<tr>
<th scope="col">
Factored form
</th>
<th scope="col">
Standard form
</th>
</tr>
</thead>
<tbody>
<tr>
<td> </td>
<td>
<p>\(5x^2+21x +4\)</p>
</td>
</tr>
</tbody>
</table>
<br>
<ol class="os-raise-noindent" type="a">
<li>This quadratic expression has a leading coefficient of 5. Identify the factors of 5 to determine the leading term
of each binomial in factored form.</li>
</ol>
<p>
\(({\style{color:red}\_}{\style{color:red}\_}{\style{color:red}\_}\;x+\_\_\_)({\style{color:red}\_}{\style{color:red}\_}{\style{color:red}\_}\;x+\_\_\_)\)
</p>
<p><strong>Answer:</strong> 5 and 1; the factored form will have the structure \((5x +\_\_\_)(x + \_\_\_)\).</p>
<ol class="os-raise-noindent" start="2" type="a">
<li>The constant term in this quadratic expression is +4. Identify all the factors of +4.</li>
</ol>
<p><strong>Answer:</strong> 1 and 4; 2 and 2</p>
<ol class="os-raise-noindent" start="3" type="a">
<li>The <span class="os-raise-ib-tooltip" data-schema-version="1.0" data-store="glossary-tooltip">linear term</span>
in the expression is \(21x\). Place each of the factors of +4 you found in part b into the remaining blanks of the
factored form structure from part a. Which of these results in a linear term of \(21x\) when you use FOIL? (Keep in
mind that the order you place these factors into your factored form structure might affect the outcome, so try both
orders if needed.)</li>
</ol>
<p><strong>Answer:</strong> 1 and 4; 1 must be placed in the binomial with \(5x\). 4 must be placed in the binomial with
\(x\).</p>
<ol class="os-raise-noindent" start="4" type="a">
<li>What is the factored form of \(5x^2+ 21x + 4\)?</li>
</ol>
<p><strong>Answer:</strong> \((5x + 1)(x + 4)\)</p>
<ol class="os-raise-noindent" start="5">
<li>Use the strategy you developed in question 4 to find the factored form of the quadratic expression shown here in
standard form.</li>
</ol>
<table class="os-raise-midsizetable">
<thead>
<tr>
<th scope="col">
Factored form
</th>
<th scope="col">
Standard form
</th>
</tr>
</thead>
<tbody>
<tr>
<td> </td>
<td>
<p>\(3x^2 + 7x + 4\)</p>
</td>
</tr>
</tbody>
</table>
<br>
<p><strong>Answer:</strong> \((3x + 4)(x + 1)\)</p>
<h4>Student Facing Extension</h4>
<h5>Are you ready for more?</h5>
<p>Find the factored form of the quadratic expression shown.</p>
<table class="os-raise-midsizetable">
<thead>
<tr>
<th scope="col">
Factored form
</th>
<th scope="col">
Standard form
</th>
</tr>
</thead>
<tbody>
<tr>
<td> </td>
<td>
<p>\(6x^2+ 11x − 10\)</p>
</td>
</tr>
</tbody>
</table>
<br>
<p>This quadratic expression is different from those in the activity, since its leading coefficient has more than one
set of possible factors.</p>
<p>Use the strategy you learned in the activity to factor the expression. If the pair of factors you selected using the
leading coefficient, 6, do not give you the correct solution, begin again with another set of factors.</p>
<p>The expression is also different since the constant term is negative, -10. This means when you consider the factors
of the constant term, one must be positive and one must be negative.</p>
<p><strong>Answer:</strong> \((3x − 2)(2x + 5)\) </p>
<h4>Anticipated Misconceptions</h4>
<p>Some students may not think to check their answers to the second question and stop as soon as they think of a pair of
factors that give the correct squared term and constant term. Encourage them to check their answers with a partner by
giving time to do so. Consider providing a non-permanent writing surface or extra paper so students could try out
their guesses and check their work without worrying about having to erase if they make a mistake the first time or
two.</p>
<h4>Activity Synthesis</h4>
<p>Select students to complete the missing expressions in standard form and to briefly explain their strategy. To
rewrite \(ax^2+ bx + c\), students are likely to have tried putting different factors of \(a\) and of \(c\) in the
factored expression such that when the factors are expanded, they yield a linear term with the coefficient \(b\).</p>
<p>Then, help students to reason about the factors more generally. Discuss questions such as:</p>
<ul>
<li> "To rewrite expressions such as \(x^2+ bx + c\), we looked for two numbers that multiply to make \(c\) and add up
to \(b\). The expressions here are of the form \(ax^2 + bx + c\). Are we still looking for two numbers that multiply
to make \(c\)? Why or why not?" (Yes. The constant terms in the factored expression must multiply to make \(c\).)
</li>
</ul>
<ul>
<li> "Do we need to look for factors of \(a\)? Why or why not?" (Yes. Those factors will be the coefficient of \(x\)
in the factored expressions. They must multiply to make \(a\).) </li>
</ul>
<ul>
<li> "Are we still looking for two factors of \(c\) that add up to \(b\)? Why or why not?" (No. The value of \(b\) is
no longer just the sum of the two factors of \(c\) because the two factors of \(a\) are now involved.) "How does
this affect the rewriting process?" (It makes it more complicated, because now there are four numbers to contend
with, and there are many more possibilities to consider.) </li>
</ul>
<p>Tell students that we'll investigate a bit further quadratic equations in the form of \(ax^2+ bx + c\) where \(a\) is
not 1, and see if there are manageable ways to rewrite such equations in factored form so that they can be solved.</p>
<h3>8.10.2: Self Check</h3>
<em>
<p class="os-raise-text-bold">After the activity, students will answer the following question to check their
understanding of the concepts explored in the activity.</p>
</em> <br>
<p class="os-raise-text-bold">QUESTION:</p>
<p>Find the factored form of the quadratic expression \(3x^2+ 5x − 2\).</p>
<table class="os-raise-textheavytable">
<thead>
<tr>
<th scope="col">
Answers
</th>
<th scope="col">
Feedback
</th>
</tr>
</thead>
<tbody>
<tr>
<td>
<p>\((3x − 2)(x − 1)\)</p>
</td>
<td>
<p>Incorrect. Let's try again a different way: This results in a middle term of \(−5x\) and a constant
term of 2. These are not correct. The answer is \((3x − 1)(x + 2)\).</p>
</td>
</tr>
<tr>
<td>
<p>\((3x + 2)(x + 1)\)</p>
</td>
<td>
<p>Incorrect. Let's try again a different way: This results in a constant term of 2. The constant term should be
−2. The answer is \((3x − 1)(x + 2)\).</p>
</td>
</tr>
<tr>
<td>
<p>\((3x − 2)(x + 1)\)</p>
</td>
<td>
<p>Incorrect. Let's try again a different way: This results in a middle term of \(3x − 2x\), or \(x\). The
middle term should be \(5x\). The answer is \((3x − 1)(x + 2)\).</p>
</td>
</tr>
<tr>
<td>
<p>\((3x − 1)(x + 2)\)</p>
</td>
<td>
<p>That's correct! Check yourself: Multiply the factored form to check to see if it is equivalent to the
original expression. \((3x −1)(x + 2) = 3x^2 +6x −x −2 = 3x^2+5x −2\). So the answer
is correct.</p>
</td>
</tr>
</tbody>
</table>
<br>
<h3>8.10.2: 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>Working with Expressions of the Form \((px + m)(qx + n)\)</h4>
<p>You have factored and solved many different types of quadratic equations, but the leading coefficient of the \(x^2\)
term was often 1.</p>
<p>Let's look at the factored form of a more complex expression to better understand the process.</p>
<p class="os-raise-text-bold">Example 1</p>
<p>Write \((4x + 1)(x + 3)\) in standard form.</p>
<p>To find the quadratic expression in standard form, use the distributive property of multiplication over addition, or
FOIL.</p>
<p>\(\begin{array}{rcl}\left(4x+1\right)\left(x+3\right)&=& 4x^2+12x+x+3\\&=& 4x^2+13x+3\end{array}\)
</p>
<p>Now, let's try to go the other direction from standard form to factored form.</p>
<p class="os-raise-text-bold">Example 2</p>
<p>Write \(2x^2+ 13x + 15\) in factored form.</p>
<p>Since the leading coefficient is 2, we know the factored form must have the structure:</p>
<p>\((2x +\_\_\_)(x + \_\_\_)\)</p>
<p>Since the constant term is +15, the two missing numbers in the factored form must have a product of 15.</p>
<p>The possibilities include:</p>
<p> +1 and +15<br>
+3 and +5<br>
-1 and -15<br>
-3 and -5</p>
<p>Since the middle term, \(13x\), is positive, we know that only the positive factors need to be considered.</p>
<p>Since the two factors are already different inside the parentheses, the order in which we place the numbers is
important. We will try both, if necessary.</p>
<p>Let's try \({\style{color:red}1}\) and \({\style{color:red}1}{\style{color:red}5}\).</p>
<p>\(\left(2x+{\style{color:red}1}{\style{color:red}5}\right)\left(x+{\style{color:red}1}\right)\)<br>
\(\begin{array}{rcl}&=&2x^2+2x+15x+15\\&=&2x^2+17x+15\end{array}\) <br> not equivalent to the original
expression</p>
<p>Try reordering the same numbers.</p>
<p>
\(\left(2x+{\style{color:red}1}\right)\left(x+{\style{color:red}1}{\style{color:red}5}\right)\)<br>\(\begin{array}{rcl}&=&2x^2+30x+x+15\\&=&2x^2+31x+15\end{array}\)<br>
not equivalent to the original expression</p>
<p>Let's try \({\style{color:blue}3}\) and \({\style{color:blue}5}\).</p>
<p>\(\left(2x+{\style{color:blue}5}\right)\left(x+{\style{color:blue}3}\right)\)<br>
\(\begin{array}{rcl}&=&2x^2+6x+5x+15\\&=&2x^2+11x+15\end{array}\)<br>
not equivalent to the original expression</p>
<p>Try reordering the same numbers.</p>
<p>\(\left(2x + {\style{color:blue}3}\right)\left(x + {\style{color:blue}5}\right)\)<br>
\(\begin{array}{rcl}&=&2x^2+ 3x + 10x + 15 \\
&=&2x^2+ 13x + 15\end{array}\)<br>
\({\style{color:red}\checkmark}\) equivalent to the original expression</p>
<p>So, the factored form of \(2x^2 + 13x + 15\) is \((2x + 3)(x + 5)\).</p>
<h4>Try It: Working with Expressions of the Form \((px + m)(qx + n)\)</h4>
<p>Find the factored form of \(3x^2− 11x − 4\).</p>
<p>Write down your answer, then select the <strong>solution</strong> button to compare your work.</p>
<p>Here is how to find the factored form:</p>
<p>The factored form will have the structure: \((3x +\_\_\_)(x +\_\_\_)\).</p>
<p>The factors of -4 are:</p>
<p>-4 and 1<br>
4 and -1<br>
-2 and 2</p>
<p>Try -4 and 1.</p>
<p>\(\left(3x-4\right)\left(x+1\right)\)<br>\(\begin{array}{rcl}&=& 3x^2-4x+3x-4\\&=&
3x^2-x-4\end{array}\)<br>not equivalent to the original expression</p>
<p>Try reordering.</p>
<p>\(\left(3x+1\right)\left(x-4\right)\)<br>\(\begin{array}{rcl}&=& 3x^2-12x+x-4\\&=&
3x^2-11x-4\end{array}\)<br>\({\style{color:red}\checkmark}\)equivalent to the original expression</p>
<p>The factored form of \(3x^2− 11x − 4\) is \((3x + 1)(x − 4)\).</p>