From Förberedande kurs i matematik 1

Revision as of 08:35, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 2.1:3b moved to Solution 2.1:3b: Robot: moved page) ← Previous diff		Current revision (08:30, 23 September 2008) (edit) (undo) Tek (Talk | contribs) m
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	If we take out the factor 5, we see that the expression can be factorized using the conjugate rule		If we take out the factor 5, we see that the expression can be factorized using the conjugate rule
-	<math> \qquad \begin{align}	+	{{Displayed math||<math>\begin{align}
	5x^2-20&=5(x^4-4)\\		5x^2-20&=5(x^4-4)\\
	&= 5(x^2-2^2)\\		&= 5(x^2-2^2)\\
-	&= 5(x+2)(x-2)	+	&= 5(x+2)(x-2)\,\textrm{.}
-	\end{align}	+	\end{align}</math>}}
-	</math>	+
-	<!--<center> [[Image:2_1_3b.gif]] </center>-->	+
-	{{NAVCONTENT_STOP}}	+

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Current revision

If we take out the factor 5, we see that the expression can be factorized using the conjugate rule

\displaystyle \begin{align} 5x^2-20&=5(x^4-4)\\ &= 5(x^2-2^2)\\ &= 5(x+2)(x-2)\,\textrm{.} \end{align}