Simplifying Complicated Fractions for Easier Calculus Differentiation

[ggolub]

Hello folks, I am studying Calculus 1 by S. Tohmpson 'Calculus made easy'.
On p. 130 I am stuck with few quizes which require to split into fractions:

(3x^2+2x+1)/[(x+2)(x^2+x+1)]
and
(x^4+1)/(x^3+1)

It asks to break each expression into sum of more simple fractions so following it differentiation would be much simpler.
Please let me know if you know how to do it.

 

[HallsofIvy]

Hello folks, I am studying Calculus 1 by S. Tohmpson 'Calculus made easy'.
On p. 130 I am stuck with few quizes which require to split into fractions:

(3x^2+2x+1)/[(x+2)(x^2+x+1)]

Since the denominator has higher degree than the numerator and x^2+ x+ 1 cannot be factored in terms of real numbers, this can be written
\frac{A}{x+ 2}+ \frac{Bx+ C}{x^2+ x+ 1}
On method of doing that is write
\frac{3x^2+ 2x+ 1}{(x+1)(x^2+ x+ 1)}= \frac{A}{x+ 2}+ \frac{Bx+ C}{x^2+ x+ 1}
multiply both sides by (x+1)(x^2+ x+ 1) to eliminate the fractions:
3x^2+ 2x+ 1= A(x^2+ x+ 1)+ (Bx+ C)(x+ 2)[/itex]<br /> Since this is to be true for all x, you can let x be any 3 numbers you want to get 3 equations to solve for A, B,a nd C. For example x=-2 is especialy nice since then you get 3(-2)^2+ 2(-2)+ 1= 12- 4+ 1= 9= A((-2)^2+ (-2)+ 1)= 3A<br /> But taking x= 0 gives<br /> 1= A+ 2C <br /> and x= 1 gives<br /> 6= 3A+ 3B+ 3C<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> and<br /> (x^4+1)/(x^3+1) </div> </div> </blockquote>Here the numerator has higher degree than the denominator so the first thing you want to do is use &quot;long division&quot;. x^3+ 1 divides into x^4+ 1 x times with -x+ 1 as remainder: <br /> \frac{x^4+ 1}{x^3+ 1}= x+ \frac{1- x}{x^3+ 1}= x+ \frac{1- x}{(x- 1)(x^2+ x+ 1)}<br /> Now find A, B, and C so that <br /> \frac{1- x}{x^2+ x+ 1}= \frac{A}{x-1}+ \frac{B}{x^2+ x+ 1}.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> It asks to break each expression into sum of more simple fractions so following it differentiation would be much simpler.<br /> Please let me know if you know how to do it. </div> </div> </blockquote> Since this has nothing to do with &quot;Differential Equations&quot;, I am moving it to &quot;Calculus and Analysis&quot;.

 

[ggolub]

Thank you so much!
Can you please refer me the book or name of topic which I can use to find this material. I want to understand the method behind it.

 

[Strants]

(x-1)(x^{2}+x+1) = x^{3} - 1, not x^{3} + 1.

Cubics are way beyond my ability to factor, but just looking at it, but I'm pretty sure x^{3} + 1 is in lowest terms. In that case, wouldn't we leave the equation in the form it is already in? Or am I missing something?

 

[HallsofIvy]

You are right. (x- 1)(x^2+ x+ 1)= x^3- 1.

What I meant to say was x^3+ 1= (x+ 1)(x^2- x+ 1)

x^n+ 1 cannot be factored (in terms of real numbers) if n is even but if n is odd, x^n+ 1= (x- 1)(x^{n-1}- x^{n-2}+ \cdot\cdot\cdot- x+ 1).