Expert: Paul Klarreich - 3/21/2006

Question
1. My name is Danielle
2. I am studying Calculus with Analytic Geometry
3. I can follow through on the first derivative of a quotient by use of the Quotient Rule and can even apply the second derivative to the quotient. Where I am having trouble is multiplying out the often very long numerator and then factoring correctly to arrive at the same answer as my book has. So I guess my only question would be how do I properly go about multiplying a numerator that looks like
(x^2-1)^2(-4)-(-4x)(2)(x^2-1)/(x^2-1)^4?
Thanks so much for your help with this Paul.
Sincerely,
Danielle

Answer
Hi, Danielle,

You wrote:

Question:  1. My name is Danielle
2. I am studying Calculus with Analytic Geometry
3. I can follow through on the first derivative of a quotient by use of the Quotient Rule and can even apply the second derivative to the quotient. Where I am having trouble is multiplying out the often very long numerator and then factoring correctly to arrive at the same answer as my book has. So I guess my only question would be how do I properly go about multiplying a numerator that looks like
(x^2-1)^2(-4)-(-4x)(2)(x^2-1)/(x^2-1)^4?
Thanks so much for your help with this Paul.
Sincerely,
Danielle
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WARNING: THE MATERIAL BELOW MAY CONTAIN FRACTIONS AND OTHER MATERIAL INAPPROPRIATE FOR CERTAIN COMPUTING SYSTEMS.  BE SURE TO VIEW IT IN A FIXED-SIZE FONT, SUCH AS COURIER.

Your example:
(x^2-1)^2(-4) - (-4x)(2)(x^2-1)
-------------------------------
   (x^2-1)^4?

is fairly typical, but there are a large variety of these, and so:

Welcome to calculus and differentiation exercises.  (Every September, when the National Football League starts its season, and a rookie quarterback comes into the game, the 300-lb defensive linemen all tackle him, grind him into the dirt(or turf) and say "Welcome to the National Football League.)

THIS subject, not algebra class, is where you REALLY learn your algebra.

Most of it is really just a matter of being careful, and perhaps learning to simplify things in sections -- meaning don't try to do the whole thing all at once.  Like your example:

(x^2-1)^2(-4) - (-4x)(2)(x^2-1)

Here is one way to do it:

Start with squaring the (x^2 -1).  You will recall the rule that (a + b)^2 is  a^2 + 2ab + b^2, so you write:

(x^2 - 1)^2 = x^4 - 2x^2 + 1

Then you do the multiplication by (-4):

-4x^4 + 8x^2 - 4

Now you set to work on the second term:

- (-4x)(2) = + 8x   (good to get those signs straight)

+8x(x^2-1) = + 8x^3 - 8x

Now you put the two pieces together:

-4x^4 + 8x^2 - 4 + 8x^3 - 8x

and put the terms in descending order:

-4x^4 + 8x^3 + 8x^2 - 8x - 4

Now as to factoring, always remove any common factor:

4( -x^4 + 2x^3 + 2x^2 - 2x - 1)
------------------------------------------
HOWEVER, we don't do things in a vacuum.  Before mechanically carrying out all that stuff, you look to see if there are any obvious common factors.  If they are removed, you will certainly have less work:

Here is your original:

(x^2-1)^2(-4) - (-4x)(2)(x^2-1)

Of course, it has a C.F. In fact, more than one.  Take out both a factor of -4 and a factor of  (x^2 - 1):

First term:  -4(x^2 - 1)(x^2 - 1)

Second term: -4(x^2 - 1)(x)(2), and don't forget this one is preceded by a minus.

Factor the whole thing:

-4(x^2 - 1)( x^2 - 1 - (2x) ) =

-4(x^2 - 1)( x^2 - 2x - 1 )

Now see if anything can be canceled:

-4(x^2 - 1)( x^2 - 2x - 1 )
----------------------------
     (x^2-1)^4


Yes, one factor of (x^2 - 1) cancels, so we have:

-4( x^2 - 2x - 1 )
--------------------
     (x^2-1)^3

And you can multiply out the top, or not, as you please.

Do I have some sure-fire tricks to make these easy?  Yes, of course.  Do a few hundred of them and you will become terrific at this.