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<h4>Working with Expressions of the Form \((px + m)(qx + n)\)</h4>
<p>You have factored and solved many different types of <span class="os-raise-ib-tooltip" data-schema-version="1.0" data-store="glossary-tooltip">quadratic equations</span>, but the leading <span class="os-raise-ib-tooltip" data-schema-version="1.0" data-store="glossary-tooltip">coefficient</span> of the \(x^2\) term was often 1.</p>
<p>Let’s look at the <span class="os-raise-ib-tooltip" data-schema-version="1.0" data-store="glossary-tooltip">factored form</span> of a more complex expression to better understand the process.</p>
<p><strong>Example 1</strong></p>
<p>Write \((4x + 1)(x + 3)\) in <span class="os-raise-ib-tooltip" data-schema-version="1.0" data-store="glossary-tooltip">standard form</span>.</p>
<p>To find the <span class="os-raise-ib-tooltip" data-schema-version="1.0" data-store="glossary-tooltip">quadratic expression</span> 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><strong>Example 2</strong></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>
\(\begin{array}{rcl}&=&2x^2+ 10x + 3x + 15 \\
&=&2x^2+ 13x + 15\end{array}\)<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>
<div class="os-raise-ib-cta" data-button-text="Solution" data-fire-event="Reveal1" data-schema-version="1.0">
<div class="os-raise-ib-cta-content">
<p>Find the factored form of \(3x^2− 11x − 4\).</p>
</div>
<div class="os-raise-ib-cta-prompt">
<p>Write down your answer, then select the <strong>solution</strong> button to compare your work.</p>
</div>
</div>
<div class="os-raise-ib-content" data-schema-version="1.0" data-wait-for-event="Reveal1">
<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>
</div>