Finding LCM of an expression x x 2 − 8 x + 7 − x x 2 − 2 x − 35 = 6 x ( x − 7 ) ( x − 1 ) ( x + 5 ) I am currently trying to figure out how to solve this problem. I know how they got the denominator but am unsure how they got 6x in the numerator? Can somebody explain this to me?

Question

Finding LCM of an expression      x                  x        2            −      8      x      +      7        −      x                  x        2            −      2      x      −      35        =            6      x              (      x      −      7      )      (      x      −      1      )      (      x      +      5      )      I am currently trying to figure out how to solve this problem. I know how they got the denominator but am unsure how they got 6x in the numerator? Can somebody explain this to me?

Kylan Simon · Accepted Answer

x                  x        2            −      8      x      +      7        −      x                  x        2            −      2      x      −      35        =      x          (      x      −      7      )      (      x      −      1      )        −      x          (      x      −      7      )      (      x      +      5      )        =            x      (      x      +      5      )      −      x      (      x      −      1      )              (      x      −      7      )      (      x      −      1      )      (      x      +      5      )        =                    x        2            +      5      x      −              x        2            +      x              (      x      −      7      )      (      x      −      1      )      (      x      +      5      )        =            6      x              (      x      −      7      )      (      x      −      1      )      (      x      +      5      )

Marco Villanueva · Accepted Answer

Factor both denominators;       x    2    −  8  x  +  7 factors as   (  x  −  1  )  (  x  −  7  ), and       x    2    −  2  x  −  35 factors as   (  x  −  7  )  (  x  +  5  ). So   (  x  −  1  )  (  x  −  7  )  (  x  +  5  ) is a multiple of both. Now, to get each fraction to have that denominator, you need to multiply by the extra bits; for example, for       x                  x        2            −      8      x      +      7      , you need to multiply the top and bottom by   x  +  5, because that&#039;s the missing part in the bottom. That gives you                     x        2            +      5      x              (      x      −      7      )      (      x      −      1      )      (      x      +      5      )      . For the other one, the missing piece is   x  −  1, so you get                     x        2            −      x              (      x      −      7      )      (      x      −      1      )      (      x      +      5      )      . Then subtracting the numerators gives you       x    2    +  5  x  −  (      x    2    −  x  )  =  6  x

Finding LCM of an expression x x 2 </msup>

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