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Dividing Polynomials


Date: 8/18/96 at 2:35:54
From: Anonymous
Subject: Dividing Polynomials

Please help me with this problem.  I know what the answer is; I just 
can't seem to get the same solution.

      (a^3 - 6a^2 + 8a)
      -----------------
              5
  --------------------------
      (2a - 4)/(10a - 40)

Thank you.


Date: 8/30/96 at 15:47:51
From: Doctor James
Subject: Re: Dividing Polynomials

Okay, your problem is:

  ( [a^3 - 6a^2 + 8a]/5 )/( [2a - 4]/[10a - 40] )

The first thing to do is to remember (a/b)/(c/d) = (a/b)*(d/c).
So we get:

  ( [a^3 - 6a^2 + 8a]/5 )*( [10a - 40]/[2a - 4] )

Factoring out a 2 from the top and bottom of the right side gives us:

  ( [a^3 - 6a^2 + 8a]/5 )*( [5a - 20]/[a - 2] )

Now, remeber that (a/b)*(d/c) = (a/c)*(d/b). So we can get:

  ( [a^3 - 6a^2 + 8a]/[a - 2] )*( [5a - 20]/5 )

But 5 goes into 5a and 20 nicely, leaving:

  ( [a^3 - 6a^2 + 8a]/[a - 2] )*(a - 4)

Factoring out an a from the top of the left side leaves us with:

  ( a*[a^2 - 6a + 8]/[a - 2] )*(a - 4)

but we can factor the a^2 - 6a + 8  to give us:

  ( a*[a - 2]*[a - 4]/[a - 2] )*(a - 4)

But the (a - 2) on the top and bottom can cancel out! So we have:

  (a*[a - 4])*(a - 4)
or:
   a*(a - 4)^2 = a^3 - 8a^2 + 16a,

which should be the answer.

-Doctor James,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Polynomials

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