MHB Partial Fraction Decomposition when denominator can't be further factored

AI Thread Summary
To approach the fraction x^2 / (x^2 + 9), recognize that the numerator and denominator have the same degree, which complicates partial fraction decomposition. Instead, rewrite the expression as 1 - (9 / (x^2 + 9)), allowing for easier integration. This method is a common technique for handling such fractions. If the goal is integration, this transformation simplifies the process significantly. Understanding the relationship between the degrees of the numerator and denominator is crucial for determining the appropriate approach.
tmt1
Messages
230
Reaction score
0
I have this fraction

$$x^2 / (x^2 + 9)$$

I'm not sure how to approach this problem since the denominator can't be further factored. What is the right approach for this type of problem?
 
Mathematics news on Phys.org
In order to do partial fractions we must have an expression that has a lower order in the numerator than in the denominator. In your case they are the same. So you would be looking for a form like [math]\frac{Ax + B}{x^2 + 9}[/math].

I would suggest looking at it like this:
[math]\frac{x^2}{x^2 + 9} = \frac{x^2 + 9 - 9}{x^2 + 9} = 1 - \frac{9}{x^2 + 9}[/math]
which is now in the desired form.

This is a fairly common trick.

-Dan
 
tmt said:
I have this fraction: $$x^2 / (x^2 + 9)$$

I'm not sure how to approach this problem since the denominator can't be further factored.
What is the right approach for this type of problem?
it depends on what you intend to do with it.

If you are trying to decompose it into Partial Fractrions, nothing can be done.

If you are trying to integrate it:

. . \int\frac{x^2}{x^2+9}dx \;=\;\int\frac{x^2+9 - 9}{x^2+9} dx

. . =\;\int\left(\frac{x^2+9}{x^2+9} - \frac{9}{x^2+9}\right)dx \;=\;\int\left(1 - \frac{9}{x^2+9}\right) dx \;\cdots\;\text{etc.}
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top