Solution 2.1:3b

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m (Robot: Automated text replacement (-[[Bild: +[[Image:))
Current revision (08:30, 23 September 2008) (edit) (undo)
m
 
(One intermediate revision not shown.)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
 
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}}
+

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}