From Förberedande kurs i matematik 1

Revision as of 08:38, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 2.1:5c moved to Solution 2.1:5c: Robot: moved page) ← Previous diff		Current revision (10:57, 23 September 2008) (edit) (undo) Tek (Talk | contribs) m
(One intermediate revision not shown.)
Line 1:		Line 1:
-	{{NAVCONTENT_START}}	+	The fraction can be further simplified if it is possible to factorize and eliminate common factors from the numerator and denominator. Both numerator and denominator are already factorized to a certain extent, but we can go further with the numerator and break it up into linear factors by using the conjugate rule
-	<center> [[Image:2_1_5c.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\begin{align}
		+	3x^{2}-12 &= 3(x^{2}-4) = 3(x+2)(x-2)\,,\\
		+	x^{2}-1 &= (x+1)(x-1) \,\textrm{.}
		+	\end{align}</math>}}
		+
		+	The whole expression is therefore equal to
		+
		+	{{Displayed math||<math>\frac{3(x+2)(x-2)(x+1)(x-1)}{(x+1)(x+2)} = 3(x-2)(x-1)\,\textrm{.}</math>}}
		+
		+	Note: One can of course expand the expression to get <math>3x^{2}-9x+6</math>
		+	as the answer.

---

Current revision

The fraction can be further simplified if it is possible to factorize and eliminate common factors from the numerator and denominator. Both numerator and denominator are already factorized to a certain extent, but we can go further with the numerator and break it up into linear factors by using the conjugate rule

\displaystyle \begin{align} 3x^{2}-12 &= 3(x^{2}-4) = 3(x+2)(x-2)\,,\\ x^{2}-1 &= (x+1)(x-1) \,\textrm{.} \end{align}

The whole expression is therefore equal to

\displaystyle \frac{3(x+2)(x-2)(x+1)(x-1)}{(x+1)(x+2)} = 3(x-2)(x-1)\,\textrm{.}

Note: One can of course expand the expression to get \displaystyle 3x^{2}-9x+6 as the answer.