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<!DOCTYPE html>
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<title>Dividing polynomials by binomials 3</title>
<script data-main="../local-only/main.js" src="../local-only/require.js"></script>
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<body>
<div class="exercise">
<div class="vars" data-ensure="B !== 0 &amp;&amp; getGCD(A, E) === 1">
<var id="A">randRangeExclude(-9, 9, [-1, 0, 1])</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>
<div class="problems">
<div id="coefficient-in-denominator">
<div class="vars">
<var id="DENOMINATOR">plus(A + "x", E)</var>
<var id="CONDITION">fractionReduce(-E, A, true)</var>
<var id="SOLUTION">plus("x", F)</var>
</div>
<p class="problem">Simplify the right-hand side of the equation below and state the condition under which the simplification is valid.</p>
<p class="question"><code>y = \dfrac{<var>A</var>x^2+<var>coefficient(B)</var>x+<var>C</var>}{<var>DENOMINATOR</var>}</code></p>
<div class="solution" data-type="multiple">
<span><code>y = </code></span>
<div class="sol" data-type="expression" data-same-form="">
<var>SOLUTION.toString()</var>
</div>
<code>x \neq \space </code><span class="sol"><var>-E/A</var></span>
</div>
</div>
<div id="no-coefficient-in-denominator">
<div class="vars">
<var id="DENOMINATOR">plus("x", F)</var>
<var id="CONDITION">-F</var>
<var id="SOLUTION">plus(A+"x", E)</var>
</div>
<p class="problem">Simplify the right-hand side of the equation below and state the condition under which the simplification is valid.</p>
<p class="question"><code>y = \dfrac{<var>A</var>x^2+<var>coefficient(B)</var>x+<var>C</var>}{<var>DENOMINATOR</var>}</code></p>
<div class="solution" data-type="multiple">
<span><code>y = </code></span>
<div class="sol" data-type="expression" data-same-form="">
<var>SOLUTION.toString()</var>
</div>
<code>x \neq \space </code><span class="sol"><var>-F</var></span>
</div>
</div>
</div>
<div class="hints">
<p>First use factoring by grouping to factor the expression in the numerator.</p>
<div>
<p>This expression is in the form <code>\blue{A}x^2 + \green{B}x + \pink{C}</code>.</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; \red{<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>\red{<var>A * C</var>}</code> and add them together.
<span data-if="A * C < 0">
Remember, since <code>\red{<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; \red{<var>A * C</var>} \\
\purple{a} + \purple{b} &amp;=&amp; \purple{<var>E</var>} + \purple{<var>A * F</var>}
&amp;=&amp; \green{<var>B</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)
+ (\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>
<div>
<p>Now factor out <code>(<var>A</var>x + <var>E</var>)</code>:</p>
<p><code>\qquad (<var>A</var>x + <var>E</var>)(x + <var>F</var>)</code></p>
</div>
<div>
<p>The original expression can therefore be written:</p>
<p><code>\qquad y = \dfrac{(<var>A</var>x + <var>E</var>)(x + <var>F</var>)}{<var>DENOMINATOR</var>}</code></p>
</div>
<div>
<p>We can divide the numerator and denominator by <code><var>DENOMINATOR</var></code>:</p>
<p><code>\qquad y =
\dfrac{\cancel{(<var>DENOMINATOR</var>)}(<var>SOLUTION</var>)}{\cancel{<var>DENOMINATOR</var>}}
= <var>SOLUTION</var>
</code></p>
</div>
<div>
<p>Because we divided by <code><var>DENOMINATOR</var></code>,</p>
<p><code>\begin{align}
<var>DENOMINATOR</var> &amp;\neq 0 \\
x &amp;\neq <var>CONDITION</var>
\end{align}
</code></p>
</div>
<div>
<p>Therefore,</p>
<p>
<code>\qquad y = <var>SOLUTION</var></code><br>
<code>\qquad x \neq <var>CONDITION</var></code>
</p>
</div>
</div>
</div>
</body>
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