Answer 2.1:3
From Förberedande kurs i matematik 1
(Difference between revisions)
			  			                                                      
		          
			 (Ny sida: {| width="100%" cellspacing="10px" |a) |width="33%" | <(x+6)(x-6)</math> |b) |width="33%"  | <math>5(x+2)(x-2)</math> |c) |width="33%" | <math>$(x+3)^2</math> |- |d) || <math>(x-5)^2</math>...)  | 
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| Line 1: | Line 1: | ||
{| width="100%" cellspacing="10px"  | {| width="100%" cellspacing="10px"  | ||
|a)  | |a)  | ||
| - | |width="33%" | <(x+6)(x-6)</math>  | + | |width="33%" | <math>(x+6)(x-6)</math>  | 
|b)  | |b)  | ||
|width="33%"  | <math>5(x+2)(x-2)</math>  | |width="33%"  | <math>5(x+2)(x-2)</math>  | ||
|c)  | |c)  | ||
| - | |width="33%" | <math>  | + | |width="33%" | <math>(x+3)^2</math>  | 
|-  | |-  | ||
|d)  | |d)  | ||
Revision as of 11:36, 31 March 2008
| a) | \displaystyle (x+6)(x-6) | b) | \displaystyle 5(x+2)(x-2) | c) | \displaystyle (x+3)^2 | 
| d) | \displaystyle (x-5)^2 | e) | \displaystyle -2x(x+3)(x-3) | f) | \displaystyle (4x+1)^2 | 
