# Thread: Partial Fractions - How to set up numerator variables

1. ## Partial Fractions - How to set up numerator variables

Hey math wizards,

I'm in calc2 and doing partial fractions. I get that the general idea of partial fractions is to split one fraction into one or more by:
1.Fixing improper fractions (if necessary)
2. factoring the denominator
3. Substitute for ABCD (as necessary)
4. plug in convenient x values to solve for variables
5. plug in solved variables to solve for leftovers

I'm probably missing something above but I am not sure exactly how to do step 3. For example I have these problem:

a) 7x^2 +3x -3 / x^3 +x^2 and b) x^2 / x^4 -42x^2- 343

would a) numerator look like Ax^2 +Bx - C? would b) look like Ax + Ax^2? What are the general rules for setting these variables up?

2. ## Re: Partial Fractions - How to set up numerator variables

Hey math wizards,

I'm in calc2 and doing partial fractions. I get that the general idea of partial fractions is to split one fraction into one or more by:
1.Fixing improper fractions (if necessary)
2. factoring the denominator
3. Substitute for ABCD (as necessary)
4. plug in convenient x values to solve for variables
5. plug in solved variables to solve for leftovers

I'm probably missing something above but I am not sure exactly how to do step 3. For example I have these problem:

a) 7x^2 +3x -3 / x^3 +x^2 and b) x^2 / x^4 -42x^2- 343

would a) numerator look like Ax^2 +Bx - C? would b) look like Ax + Ax^2? What are the general rules for setting these variables up?
You can get help here.

3. ## Re: Partial Fractions - How to set up numerator variables

First, please, please, please use parentheses! What you wrote is, properly, $7x^2+ 3x- \frac{3x}{x^3}+ x^2$ and $\frac{x^2}{x^4}- 42x^2+ 343$ when, I am sure, you meant (7x^2 +3x -3)/ (x^3 +x^2)= $\frac{7x^2+ 3x- 3}{x^3+ x^2}$ and x^2/(x^4 -42x^2- 343)= $\frac{x^2}{x^4- 42x^2- 343}$.

I don't know what you mean by "would a) numerator look like Ax^2 +Bx - C? would b) look like Ax + Ax^2? What are the general rules for setting these variables up?". The numerator in (a) is of the form Ax^2+ Bx- C with A= 7, B= 3, and C= 3. In (b) the numerator is of the form Ax+ Bx^2 with A= 0 and B= 1.

But that really isn't important. The key point in "partial fractions" is the denominator. Any polynomial can be factored into linear factors and quadratic factors with real coefficients. The first denominator, $x^3+ x^2$, can be factored as $x^2(x+ 1)= x(x)(x+ 1)$. Any fraction with that denominator can be written as $\frac{A}{x}+ \frac{B}{x^2}+ \frac{C}{x+ 1}$ with constant A, B, and C. To write $\frac{7x^2+ 3x- 3}{x^2(x+ 1)}= \frac{A}{x}+ \frac{B}{x^2}+ \frac{C}{x+ 1}$, multiply both sides by $x^2(x+ 1)$ to get $7x^2+ 3x- 3= Ax(x+ 1)+ B(x+ 1)+ Cx^2$. To determine A, B, and C, take three values for x to get three equations to solve for A, B, and C. In particular, if we let x= 0 we get -3= B and if we let x= -1, we get $7- 3- 3= 1= C$. For a third equation, set x= 1 just because that is a simple number: $7+ 3- 3= 7= 2A+ 2B+ C= 2A- 6+ 1= 2A- 5$. $2A= 7+ 5= 12$ so $A= 6$. That is $\frac{7x^2+ 3x- 3}{x^2(x+ 1)}= \frac{6}{x}- \frac{3}{x^2}+ \frac{1}{x+ 1}$.

The second problem is a bit more complicated. The denominator is fourth degree but has only even powers so let $y= x^2$. Then the denominator is the quadratic $y^2- 42y- 343$ and, if the factors are not obvious (they weren't to me!) use the quadratic formula: $y= \frac{42\pm\sqrt{42^2+ 4(343)}}{2}= \frac{42\pm\sqrt{3136}}{2}= \frac{42\pm 56}{2}$ so that $y= \frac{98}{2}= 49$ and $y= \frac{-14}{2}= -7$. With y= 49, x= 7 and x= -7 while $y= x^2= -7$ we have $x^2+ 7$ as a factor.

We can write $\frac{x^2}{x^4- 42x^2- 343}= \frac{A}{x- 7}+ \frac{B}{x+ 7}+ \frac{Cx+ D}{x^2+ 7}$. Determine A, B, C, and D by taking four different values for x. x= 7 and x= -7 give simple equations. Taking x to be the simple values, x= 0 and x= 1, give two other equations.