Solution 2.1:3f
From Förberedande kurs i matematik 1
(Difference between revisions)
			  			                                                      
		          
			 (Ny sida: {{NAVCONTENT_START}} <center> Bild:2_1_3f.gif </center> {{NAVCONTENT_STOP}})  | 
				|||
| Line 1: | Line 1: | ||
{{NAVCONTENT_START}}  | {{NAVCONTENT_START}}  | ||
| - | <center> [[Bild:2_1_3f.gif]] </center>  | + | Treating <math>4x</math> as one term, we can write  | 
| + | |||
| + | <math> \qquad 16x^2+8x+1=(4x)^2 +2\cdot 4x +1 </math>  | ||
| + | |||
| + | and because <math> y^2 +2y+1=(y+1)^2 </math> we obtain  | ||
| + | |||
| + | <math> \qquad (4x)^2 +2\cdot 4x +1= (4x+1)^2 </math>  | ||
| + | <!--<center> [[Bild:2_1_3f.gif]] </center>-->  | ||
{{NAVCONTENT_STOP}}  | {{NAVCONTENT_STOP}}  | ||
Revision as of 13:23, 13 August 2008
Treating \displaystyle 4x as one term, we can write
\displaystyle \qquad 16x^2+8x+1=(4x)^2 +2\cdot 4x +1
and because \displaystyle y^2 +2y+1=(y+1)^2 we obtain
\displaystyle \qquad (4x)^2 +2\cdot 4x +1= (4x+1)^2
