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      <text>1.3.5</text>
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      <text><![CDATA[<h3>5. Factorising quadratic expressions</h3>
<p>Quadratic expressions commonly occur in many areas of mathematics, physics and engineering. Many
quadratic expressions can be written as the product of two linear factors and, in this Section, we
examine how these factors can be easily found:</p>

<div class="HELM_keypoint">
<h4>Key Point&nbsp;</h4>
An expression of the form
\[ ax^2 + bx + c, \hspace{1cm} a \neq 0, \]
where \(a\), \(b\) and \(c\) are numbers is called a <b>quadratic expression</b> (in the variable x).<p></p>
</div>
<p><strong> </strong></p>
<p>The numbers \(b\) and \(c\) may be zero but a must not be zero (for, then, the quadratic reduces to a linear expression or constant). The number a is called the <b>coefficient</b> of \(x^2\), \(b\) is the &nbsp;coefficient&nbsp;of \(x\) and \(c\) is called the <b>constant term</b>.</p><p><br></p><b>

Case 1</b>&nbsp;<p>Consider the product \( (x+1)(x+2)\). Removing brackets yields \( x^2 +3x+2 \). Conversely, we see that
the factors of \( x^2+3x+2\) are \( (x+1)\) and \( (x+2)\). However, if we were given the quadratic expression first, how would we factorise it? The following examples show how to do this but note that not all
quadratic expressions can be easily factorised.

To enable us to factorise a quadratic expression in which the coefficient of \(x^2\^ equals 1, we note the
following expansion:
\[ (x + m)(x + n) = x^2 + mx + nx + mn = x^2 + (m + n)x + mn \]
So, given a quadratic expression we can think of the coefficient of \(x\) as \(m+n\) and the constant term
as \(mn\). Once the values of \(m\) and \(n\) have been found the factors can be easily obtained.

</p><hr>
<h4 class="HELM_example">Example</h4>
<p>Factorise \(x^2 + 4x - 5\).</p>

<h4 class="HELM_solution">Solution</h4><p>Writing \(x^2 + 4x - 5 = (x + m)(x + n) = x2 + (m + n)x + mn \), we seek numbers \(m\) and \(n\) such that \(m+ n = 4\) and \(mn = -5\). By trial and error it is not difficult to find that \(m = 5\) and \(n = -1 \) (or, the other way round, \( m = -1 \) and \(n = 5\) ). So we can write&nbsp;</p><p>\[ x2 + 4x - 5 = (x + 5)(x - 1) \] The answer can be checked easily by removing brackets.</p><b><p><b><br></b></p>

Case 2</b>&nbsp;<p>When the coefficient of \(x^2\) is not equal to \(1\) it may be possible to extract a numerical factor. For
example, note that \(3x^2 + 18x + 24\) can be written as \(3(x^2 + 6x + 8)\) and then factorised as in
the previous Example 56. Sometimes no numerical factor can be found and a slightly different
approach may be taken. We will demonstrate a technique which can always be used to transform
the given expression into one in which the coefficient of the squared variable equals 1.<br></p><p></p><hr>
<h4 class="HELM_example">Example</h4>
<p>Factorise \(2x^2 + 5x + 3\).</p>

<h4 class="HELM_solution">Solution</h4><p>
First note the coefficient of \(x^2\); in this case is \(2\). Multiply the whole expression by this number and
rearrange as follows:
\[ 2(2x^2 + 5x + 3) = 2(2x^2) + 2(5x) + 2(3) = (2x^)2 + 5(2x) + 6. \]
We now introduce a new variable \(z\) such that \(z = 2x\) Thus we can write
\[ (2x)^2 + 5(2x) + 6 \text{  as  } z^2 + 5z + 6. \]
This can be factorised to give \( (z + 3)(z + 2)\). Returning to the original variable by replacing \(z\) by
\(2x\) we find
\[ 2(2x^2 + 5x + 3) = (2x + 3)(2x + 2) \]
A factor of \(2\) can be extracted from the second bracket on the right so that
\[ 2(2x^2 + 5x + 3) = 2(2x + 3)(x + 1), \]
so that
\[ 2x^2 + 5x + 3 = (2x + 3)(x + 1). \]</p>

As an alternative to the technique of Example 56, experience and practice can often help us to
identify factors. For example suppose we wish to factorise \(3x^2 + 7x + 2\). We write
\[ 3x^2 + 7x + 2 = ( \hspace{1cm} )( \hspace{1cm} ) \]
In order to obtain the term \(3x^2\) we can place terms \(3x\) and \(x\) in the brackets to give
\[ 3x^2 + 7x + 2 = (3x + ? )(x + ? ) \]

In order to obtain the constant \(2\), we consider the factors of \(2\). These are \(1\),\(2\) or \(-1\),\(-2\). By placing these factors in the brackets we can factorise the quadratic expression. Various possibilities exist: we could write \( (3x+2)(x+1)\) or \((3x+1)(x+2)\) or \((3x - 2)(x - 1)\) or \( (3x - 1)(x -2)\), only one of which
is correct. By removing brackets from each in turn we look for the factorisation which produces the
correct middle term, \(7x\). The correct factorisation is found to be
\[ 3x^2 + 7x + 2 = (3x + 1)(x + 2). \]
With practice you will be able to carry out this process quite easily.<p></p><p></p>

<h4 class="HELM_example">Example</h4>
<p>Factorise</p>
<ol class="HELM_parts_inline">
<li>\( x^2 - 36z^2,\)</li>
<li>\(25x^2 - 9z^2,\)</li>
<li>\(\alpha^2 - 1\)</li>
</ol>

<h4 class="HELM_solution">Solution</h4>
In each case we are required to find the difference of two squared terms.<p>

</p><ol class="HELM_parts">
  <li>Note that \(x^2 - 36z^2 = x^2 - (6^z)2.\) This factorises as \( (x + 6z)(x - 6z).\)</li>
  <li>Here \(25x^2 - 9z^2 = (5x)^2 - (3z)^2.\) This factorises as \((5x + 3z)(5x - 3z).\)</li>
  <li>\( \alpha^2 - 1 = (\alpha + 1)(\alpha - 1). \)<p></p></li></ol>]]></text>
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