From Förberedande kurs i matematik 1

Revision as of 08:36, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 2.1:3e moved to Solution 2.1:3e: Robot: moved page) ← Previous diff		Current revision (08:37, 23 September 2008) (edit) (undo) Tek (Talk | contribs) m
Line 1:		Line 1:
-	{{NAVCONTENT_START}}	+	Both terms contain ''x'', which can therefore be taken out as a factor (as can 2),
-	<!--center> [[Image:2_1_3e.gif]] </center-->	+
-	Both terms contain <math>x</math>, which can therefore be taken out as a factor (as can 2).	+
-	:<math>18x-2x^3=2x\cdot 9-2x \cdot x^2=2x(9-x). </math>	+	{{Displayed math||<math>18x-2x^3=2x\cdot 9-2x \cdot x^2=2x(9-x^2)\,\textrm{.}</math>}}
	The remaining second-degree factor <math> 9-x^2 </math> can then be factorized using the conjugate rule		The remaining second-degree factor <math> 9-x^2 </math> can then be factorized using the conjugate rule
-	:<math> 2x(9-x^2)=2x(3^2-x^2)=2x(3+x)(3-x), </math>	+	{{Displayed math||<math> 2x(9-x^2)=2x(3^2-x^2)=2x(3+x)(3-x)\,,</math>}}
-	which can also be written as <math> -2x(x+3)(x-3).</math>	+	which can also be written as <math>-2x(x+3)(x-3).</math>
-	{{NAVCONTENT_STOP}}	+

---

Current revision

Both terms contain x, which can therefore be taken out as a factor (as can 2),

\displaystyle 18x-2x^3=2x\cdot 9-2x \cdot x^2=2x(9-x^2)\,\textrm{.}

The remaining second-degree factor \displaystyle 9-x^2 can then be factorized using the conjugate rule

\displaystyle 2x(9-x^2)=2x(3^2-x^2)=2x(3+x)(3-x)\,,

which can also be written as \displaystyle -2x(x+3)(x-3).