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October 10th, 2014, 05:43 AM  #1 
Senior Member Joined: Aug 2014 From: Mars Posts: 101 Thanks: 9  LCD and prime numbers(with binomials)
So I was given this problem............. Find the LCD= $\displaystyle \frac{3}{2x6}, \frac{4}{x^29}, \frac{18}{6x+18}$ The break down (do I call these the prime numbers? $\displaystyle 2(x3), (x3)(x+3), 6(x+3))$ My answer: $\displaystyle 6(x+3) \cdot 2(x3)$ Their answer: $\displaystyle 6(x^29)$????? what? It seems they disregarded the 2, but why? Help me understand please. Last edited by skipjack; October 11th, 2014 at 04:04 PM. 
October 10th, 2014, 06:33 AM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond 
$\displaystyle 2(x3)\cdot3(x+3)=6(x^29)$ (The least common multiple of 2 and 6 is 6). They are not necessarily prime numbers. Last edited by greg1313; October 10th, 2014 at 06:35 AM. 
October 10th, 2014, 08:21 PM  #3  
Senior Member Joined: Aug 2014 From: Mars Posts: 101 Thanks: 9  Quote:
On that same note if I had $\displaystyle 4(x3), 6(x+3)$ would I some how choose $\displaystyle 12(x^29)$ I don't quite understand this.  
October 10th, 2014, 08:43 PM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond 
That's correct. You're looking for the least common denominator  the least common multiple of 4 and 6 is 12. So if you had $\displaystyle \frac{3}{4(x3)}+\frac{3}{6(x+3)}$ you'd write it as $\displaystyle \frac{3}{4(x3)}\cdot\frac{3(x+3)}{3(x+3)}+\frac{3}{6(x+3)} \cdot\frac{2(x3)}{2(x3)}$ $\displaystyle =\frac{9(x+3)+6(x3)}{12(x3)(x+3)}$ $\displaystyle =\frac{15x+9}{12(x3)(x+3)}$ $\displaystyle \frac{3(5x+3)}{12(x3)(x+3)}$ $\displaystyle =\frac{5x+3}{4(x3)(x+3)}$ Last edited by greg1313; October 10th, 2014 at 08:50 PM. 
October 10th, 2014, 09:59 PM  #5 
Senior Member Joined: Apr 2014 From: Europa Posts: 584 Thanks: 177  $\displaystyle \color{green}{2x6=2(x3) \\\;\\ x^29=x^23^2=(x3)(x+3) \\\;\\ 6x+18=6(x+3)=2\cdot3(x+3) \\\;\\ LCD \\\;\\ 2\cdot3(x+3)(x3)=6(x^29)}$ Last edited by skipjack; October 11th, 2014 at 04:05 PM. 

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