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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= $\displaystyle \frac{3}{2x-6}, \frac{4}{x^2-9}, \frac{18}{6x+18}$ The break down (do I call these the prime numbers? $\displaystyle 2(x-3), (x-3)(x+3), 6(x+3))$ My answer: $\displaystyle 6(x+3) \cdot 2(x-3)$ Their answer: $\displaystyle 6(x^2-9)$????? 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 $\displaystyle 2(x-3)\cdot3(x+3)=6(x^2-9)$ (The least common multiple of 2 and 6 is 6). They are not necessarily prime numbers. Thanks from Opposite Last edited by greg1313; October 10th, 2014 at 06:35 AM. October 10th, 2014, 08:21 PM   #3
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Quote:
 Originally Posted by greg1313 $\displaystyle 2(x-3)\cdot3(x+3)=6(x^2-9)$ (The least common multiple of 2 and 6 is 6). They are not necessarily prime numbers.

On that same note if I had $\displaystyle 4(x-3), 6(x+3)$ would I some how choose $\displaystyle 12(x^2-9)$ 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 $\displaystyle \frac{3}{4(x-3)}+\frac{3}{6(x+3)}$ you'd write it as $\displaystyle \frac{3}{4(x-3)}\cdot\frac{3(x+3)}{3(x+3)}+\frac{3}{6(x+3)} \cdot\frac{2(x-3)}{2(x-3)}$ $\displaystyle =\frac{9(x+3)+6(x-3)}{12(x-3)(x+3)}$ $\displaystyle =\frac{15x+9}{12(x-3)(x+3)}$ $\displaystyle \frac{3(5x+3)}{12(x-3)(x+3)}$ $\displaystyle =\frac{5x+3}{4(x-3)(x+3)}$ Thanks from Opposite Last edited by greg1313; October 10th, 2014 at 08:50 PM. October 10th, 2014, 09:59 PM   #5
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Quote:
 Originally Posted by Opposite Their answer: $\displaystyle 6(x^2-9)$????? what?
$\displaystyle \color{green}{2x-6=2(x-3) \\\;\\ x^2-9=x^2-3^2=(x-3)(x+3) \\\;\\ 6x+18=6(x+3)=2\cdot3(x+3) \\\;\\ LCD \\\;\\ 2\cdot3(x+3)(x-3)=6(x^2-9)}$

Last edited by skipjack; October 11th, 2014 at 04:05 PM. Tags binomials, lcd, numberswith, prime lcd of binomials

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