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		<title>1.3 Lösning 6c - Versionshistorik</title>
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		<updated>2026-11-06T01:32:40Z</updated>
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	<entry>
		<id>https://wiki.mattekollen.se/index.php?title=1.3_L%C3%B6sning_6c&amp;diff=2438&amp;oldid=prev</id>
		<title>Taifun den 20 februari 2011 kl. 20.20</title>
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				<updated>2011-02-20T20:20:27Z</updated>
		
		<summary type="html">&lt;p&gt;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;Ny sida&lt;/b&gt;&lt;/p&gt;&lt;div&gt;För att faktorisera polynomet &amp;lt;math&amp;gt; 4\,x^2 - 36 &amp;lt;/math&amp;gt; beräknar vi dess nollställen: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\begin{align} 4\,x^2 - 36 &amp;amp; = 0   &amp;amp; &amp;amp; | \;\;\; / \,4     \\&lt;br /&gt;
                       x^2 -  9 &amp;amp; = 0   &amp;amp; &amp;amp; | \;\; +9          \\&lt;br /&gt;
                       x^2      &amp;amp; =  9  &amp;amp; &amp;amp; | \;\; \sqrt{\;\;} \\&lt;br /&gt;
                       x_1      &amp;amp; = 3                          \\&lt;br /&gt;
                       x_2      &amp;amp; = - 3                        \\&lt;br /&gt;
     \end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Därför har polynomet &amp;lt;math&amp;gt; x^2 - 9\, &amp;lt;/math&amp;gt; faktorformen: &amp;lt;math&amp;gt; (x+3) \cdot (x-3) &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
Det ursprungliga polynomet &amp;lt;math&amp;gt; 4\,x^2 - 36 &amp;lt;/math&amp;gt; har faktorformen: &amp;lt;math&amp;gt; 4\cdot (x+3) \cdot (x-3) &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
Kontroll:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 4\cdot (x+3) \cdot (x-3) = 4 \cdot (x^2 - 9) = 4\,x^2 - 36 &amp;lt;/math&amp;gt;&lt;/div&gt;</summary>
		<author><name>Taifun</name></author>	</entry>

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