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      <text>1.1.11 (Factorising coefficient not 1)</text>
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      <text><![CDATA[<h1>Coefficient of \(x^2\) is not \(1\)</h1>

<p>It is possible to factorise some quadratics where the leading coefficient is not equal to one. Whilst the method to factorise such quadratics is similar to the case with a leading coefficient of \(1\) it is often more difficult to do by inspection. It can often take more effort to spot the required factors. Let us consider the following example in order to explain this process. </p>
<br>
<h3>Example</h3>

<p>Factorise and solve the quadratic \(3x^2 + 5x +2 = 0\).</p>

<h3>Solution</h3>

<p>Proceed similarly to before. As mentioned previously, for the quadratic \(ax^2 + bx+c = 0\) we need to find two numbers that multiply to get \(ac\) and sum to get \(b\). So, for this example we require numbers \(m\) and \(n\) such that 
\[
mn = ac = 6 \qquad \mbox{ and } \qquad m+ n = 5.
\]
By inspection or otherwise we can see that the values which work are 2 and 3.
Therefore we have 
\[
\begin{eqnarray*}
3x^2 + 5x +2 &amp;= 3x^2 +2x + 3x + 2
\\
&amp;= x(3x+ 2) + (3x + 2)
\\
&amp;= (3x+ 2)(x+1) = 0.
\end{eqnarray*}
\]
Therefore by the null factor law, the roots of the quadratic are the values of \(x\) for which either the first or second bracket is equal to zero. Hence the solutions of the quadratic are \(x = -2/3\) or \(x = -1\)</p>

<p>Let us check that these are indeed roots of the quadratic. 
\[
-2/3: \:\:\: 3(\frac{-2}{3})^2 + (5 \cdot \frac{-2}{3}) + 2 = \frac{4}{3} - \frac{10}{3} + 2 = 0 \qquad \mbox{ and } \qquad -1: \:\:\: 3(-1)^2 + (5 \cdot -1) + 2 = 3 - 5 + 2 = 0.
\]
Hence \(-1\) and \(-\frac{2}{3}\) are roots of the given equation. Notice that in this case we can have fractional roots. 
</p>]]></text>
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