fractions with polynomial denominators

B

blubelle

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I hope I am translating this from paper to computer in a readable way... 1st timer!
I am trying to write the following in simplest form...
((a^2+1)/(a^2-1))-(a/(a+1)

I get an LCD of (a^2-1)(a+1),

then ((a^3+a^2+a+1)/(a^2-1)(a+1))-((a^3-a)/((a^2-1)(a+1))

then (a^2+1)/((a^2-1)(a+1))

then -(a^2-1)/((a^2-1)(a+1))

then -1/(a+1)


The book I am working from says the answer is 1/(a-1)... where am I going wrong? Thanks!
 
\(\displaystyle \L \frac{a^2+1}{a^2-1}\,-\, \frac{a}{a+1}\) =

Common denominator:


=\(\displaystyle \L \frac{(a+1)(a^2+1) - a(a^2-1)}{(a+1)(a^2-1)}\)

FOIL:

=\(\displaystyle \L \frac{a^3+a^2+a+1\,\,-\,\,a^3 + a}{a^3+a^2-a-1}\)

Combine like terms:

=\(\displaystyle \L \frac{a^2+2a+1}{a^3+a^2-a-1}\)

Factor:

=\(\displaystyle \L \frac{(a+1)(a+1)}{(a-1)(a+1)(a+1)}\)

Now cancel out. See?



John.
 
blubelle said:
I hope I am translating this from paper to computer in a readable way... 1st timer!
I am trying to write the following in simplest form...
((a^2+1)/(a^2-1))-(a/(a+1)

I get an LCD of (a^2-1)(a+1),

then ((a^3+a^2+a+1)/(a^2-1)(a+1))-((a^3-a)/((a^2-1)(a+1))

then (a^2+1)/((a^2-1)(a+1))

then -(a^2-1)/((a^2-1)(a+1))

then -1/(a+1)


The book I am working from says the answer is 1/(a-1)... where am I going wrong? Thanks!
Click to expand...

(a<SUP>2</SUP> - 1)(a + 1) is A common denominator, but it is not the LEAST common denominator.

I strongly suggest that you factor each of the original denominators, and use the factored form to find the LCD.

Code: 
a^2 + 1                a
--------------   -   ------       
a^2 - 1               (a + 1)

Factor the denominator of the first fraction:

  a^2 + 1               a
-------------   -    --------
(a+1)(a - 1)         (a + 1)

Now....do you see that (a + 1)(a - 1) is the LCD?

The first fraction already HAS that denominator.  Multiply numerator and denominator of the second fraction by (a - 1):

   a^2 + 1           a(a - 1)
-------------   -   ----------------
(a+1)(a - 1)      (a + 1)(a - 1)

Now...can you take it from here?
 
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