Hello, I am studying for math and I am prepared for the test because I have memo

Question

Hello, I am studying for math and I am prepared for the test because I have memorized information but now I am determined to understand reasons behind doing processes that I have memorized. I am becoming an engineer so it's in my nature to understand everything about what I am memorizing. I have two questions about partial fractions.

1.

Why is it you must add a Bx+C over top of the (x^2+1)? Why is that Bx needed?

2.

Why must you write out a fraction for each power when the power of the paranthesis is higher than 1?

I'm hoping for a in depth understanding of these two conecpts. Any help would be appreciated, thanks in advance!

Explanation / Answer

1) When you have a quadratic factor you need to include this partial fraction:

B1x + C1

(Your Quadratic)

Factors with Exponents

Sometimes you may get a factor with an exponent, like (x-2)3 ...

You need a partial fraction for each exponent from 1 up.

Like this:

Example:

1

(x-2)3

Will have partial fractions

A1 + A2 + A3

x-2 (x-2)2 (x-2)3

The same thing can also happen to quadratics:

Example:

1

(x2+2x+3)2

Will have partial fractions:

B1x + C1 + B2x + C2

x2+2x+3 (x2+2x+3)2

Sometimes Using Roots Does Not Solve It

Even after using the roots (zeros) of the bottom you can end up with unknown constants.

So the next thing to do is:

Gather all powers of x together and then solve it as a system of linear equations.

Oh my gosh! That is a lot to handle! So, on with an example to help you understand:

A Big Example Bringing It All Together

Here is a nice example for you!

x2+15

(x+3)2 (x2+3)

Because (x+3)2 has an exponent of 2, it needs two terms (A1 and A2).

And (x2+3) is a quadratic, so it will need Bx + C:

x2+15 = A1 + A2 + Bx + C

(x+3)2(x2+3) x+3 (x+3)2 x2+3

Now multiply through by (x+3)2(x2+3):

x2+15 = (x+3)(x2+3)A1 + (x2+3)A2 + (x+3)2(Bx + C)

There is a zero at x = -3 (because x+3=0), so let us try that:

(-3)2+15 = 0 + ((-3)2+3)A2 + 0

And simplify it to:

24 = 12A2

so A2=2

Let us replace A2 with 2:

x2+15 = (x+3)(x2+3)A1 + 2x2+6 + (x+3)2(Bx + C)

Now expand the whole thing:

x2+15 = (x3+3x+3x2+9)A1 + 2x2+6 + (x3+6x2+9x)B + (x2+6x+9)C

Gather powers of x together:

x2+15 = x3(A1+B)+x2(3A1+6B+C+2)+x(3A1+9B+6C)+(9A1+6+9C)

Separate the powers and write as a Systems of Linear Equations:

x3: 0 = A1+B

x2: 1 = 3A1+6B+C+2

x: 0 = 3A1+9B+6C

Constants: 15 = 9A1+6+9C

Simplify, and arrange neatly:

0 = A1 + B

-1 = 3A1 + 6B + C

0 = 3A1 + 9B + 6C

1 = A1 + C

Now solve.

You can choose your own way to solve this ... I decided to subtract the 4th equation from the 2nd to begin with:

0 = A1 + B

-2 = 2A1 + 6B

0 = 3A1 + 9B + 6C

1 = A1 + C

Then subtract 2 times the 1st equation from the 2nd:

0 = A1 + B

-2 = 4B

0 = 3A1 + 9B + 6C

1 = A1 + C

Now I know that B = -(1/2).

And from the 1st equation I can figure that A1 = +(1/2).

And from the 1st equation I can figure that C = +(1/2).

Final Result:

A1=1/2 A2=2 B=-(1/2) C=1/2

And we can now write our partial fractions:

x2+15 = 1 + 2 + -x+1

(x+3)2(x2+3) 2(x+3) (x+3)2 2(x2+3)

Phew! Lots of work. But it can be done.

2)When you have a quadratic factor you need to include this partial fraction:

B1x + C1

(Your Quadratic)

Factors with Exponents

Sometimes you may get a factor with an exponent, like (x-2)3 ...

You need a partial fraction for each exponent from 1 up.

Like this:

Example:

1

(x-2)3

Will have partial fractions

A1 + A2 + A3

x-2 (x-2)2 (x-2)3

The same thing can also happen to quadratics:

Example:

1

(x2+2x+3)2

Will have partial fractions:

B1x + C1 + B2x + C2

x2+2x+3 (x2+2x+3)2

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