Processing Math: Done

From Förberedande kurs i matematik 1

Revision as of 11:07, 31 March 2008 (edit) Lina (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:2_1_5c.gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (10:57, 23 September 2008) (edit) (undo) Tek (Talk | contribs) m
(3 intermediate revisions 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> [[Bild: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

3x2−12x2−1=3(x2−4)=3(x+2)(x−2)=(x+1)(x−1).

The whole expression is therefore equal to

(x+1)(x+2)3(x+2)(x−2)(x+1)(x−1)=3(x−2)(x−1).

Note: One can of course expand the expression to get 3x2−9x+6 as the answer.