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<!DOCTYPE html>
<html data-require="math math-format graphie">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>Factoring polynomials by grouping</title>
<script data-main="../local-only/main.js" src="../local-only/require.js"></script>
</head>
<body>
<div class="exercise">
<div class="problems">
<div>
<div class="vars" data-ensure="B !== 0 &amp;&amp; getGCD(A, E) === 1">
<var id="A" data-ensure="abs(A) !== 1">randRangeNonZero(-9, 9)</var>
<var id="E">randRangeNonZero(-9, 9)</var>
<var id="F">randRangeNonZero(-5, 5)</var>
<var id="B">E + A * F</var>
<var id="C">E * F</var>
</div>
<p class="question">
Factor the expression below completely. All coefficients should be integers.
</p>
<p class="problem">
<span id="question-a"><code><var>A</var></code></span><code>x^2<span data-if="B > 0">+</span></code>
<span id="question-b"><code><var>B</var></code></span><code>x<span data-if="C > 0">+</span></code>
<span id="question-c"><code><var>C</var></code></span>
</p>
<p class="solution" data-type="expression" data-same-form="">
(<var>A</var>x + <var>E</var>)(x + <var>F</var>)
</p>
<div class="hints">
<div>
<p>
This expression is in the form <code>\blue{A}x^2 + \green{B}x + \pink{C}</code>.
You can factor it by grouping.
</p>
<div class="graphie" style="outline: none;">
$("#question-a").addClass("hint_blue");
$("#question-b").addClass("hint_green");
$("#question-c").addClass("hint_pink");
</div>
</div>
<div>
<p>
First, find two values, <code class="hint_purple">a</code> and
<code class="hint_purple">b</code>, so:
</p>
<code>
\qquad \begin{eqnarray}
\purple{ab} &amp;=&amp; \blue{A}\pink{C} \\
\purple{a} + \purple{b} &amp;=&amp; \green{B}
\end{eqnarray}
</code>
</div>
<div>
<p>In this case:</p>
<code>
\qquad \begin{eqnarray}
\purple{ab} &amp;=&amp;
\blue{(<var>A</var>)}\pink{(<var>C</var>)} &amp;=&amp; <var>A * C</var> \\
\purple{a} + \purple{b} &amp;=&amp; &amp; &amp;
\green{<var>B</var>}
\end{eqnarray}
</code>
</div>
<p>
In order to find <code>\purple{a}</code> and <code>\purple{b}</code>, list out the factors of
<code><var>A * C</var></code> and add them together.
<span data-if="A * C < 0">
Remember, since <code><var>A * C</var></code> is negative, one of the factors must be
negative.
</span>
The factors that add up to <code>\green{<var>B</var>}</code> will be your
<code>\purple{a}</code> and <code>\purple{b}</code>.
</p>
<div>
<p>
When <code>\purple{a}</code> is <code>\purple{<var>E</var>}</code> and
<code>\purple{b}</code> is <code>\purple{<var>A * F</var>}</code>:
</p>
<code>
\qquad \begin{eqnarray}
\purple{ab} &amp;=&amp; (\purple{<var>E</var>})(\purple{<var>A * F</var>})
&amp;=&amp; <var>E * A * F</var> \\
\purple{a} + \purple{b} &amp;=&amp; \purple{<var>E</var>} + \purple{<var>A * F</var>}
&amp;=&amp; <var>E + A * F</var>
\end{eqnarray}
</code>
</div>
<div>
<p>
Next, rewrite the expression as <code>\blue{A}x^2 + \purple{a}x + \purple{b}x + \pink{C}</code>:
</p>
<code>
\qquad \blue{<var>A</var>}x^2
<span data-if="E > 0">+</span>\purple{<var>E</var>}x
<span data-if="A * F > 0">+</span>\purple{<var>A * F</var>}x
<span data-if="C > 0">+</span>\pink{<var>C</var>}
</code>
</div>
<div>
<p>
Group the terms so that there is a common factor in each group:
</p>
<code>
\qquad (\blue{<var>A</var>}x^2 <span data-if="E > 0">+</span>\purple{<var>E</var>}x)
+ (\purple{<var>A * F</var>}x <span data-if="C > 0">+</span>\pink{<var>C</var>})
</code>
</div>
<div>
<p>
Factor out the common factors:
</p>
<code>
\qquad x(<var>A</var>x + <var>E</var>) + <var>F</var>(<var>A</var>x + <var>E</var>)
</code>
</div>
<p>
Notice how <code>(<var>A</var>x + <var>E</var>)</code> has become a common factor.
Factor this out to find the answer.
</p>
<p><code>(<var>A</var>x + <var>E</var>)(x + <var>F</var>)</code></p>
</div>
</div>
</div>
</div>
</body>
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