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Partial fractions

Can anybody resolve the following into partial fractions and show me how to go about doing so?

6x / x^4-16

and

x^3 / x^2-3x+2

Thanks

Reply 1

BabyMaths

Original post by JordanSHowarth

Can anybody resolve the following into partial fractions and show me how to go about doing so?

6x / x^4-16

and

x^3 / x^2-3x+2

Thanks

First question: As a first step you should factorise.

Second: This is an improper rational expression. You could use polynomial division.

Reply 2

JordanSHowarth

The two answers that I have arrived at are:

1) 6x / x^4-16 = (3/2)/(x^2-4) +(3/2)/(x^2+4) - doesn't seem right to me, with a fraction over the fraction?

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

Reply 3

brianeverit

Original post by JordanSHowarth

The two answers that I have arrived at are:

1) 6x / x^4-16 = (3/2)/(x^2-4) +(3/2)/(x^2+4) - doesn't seem right to me, with a fraction over the fraction?

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

It is perfectly acceptable to have a fraction in the numerator since, for example

\frac{\frac{3}{2}}{x^2+4}\text{ is the same as }\frac{3}{2(x^2+4)}

However your answer is incorrect. Since

x^2-16=(x^2+4)(x^2-4)=(x^2+4)(x+2)(x-2)

you should be finding fractions of the form

\frac{A}{x+2}+\frac{B}{x-2}+\frac{Cx+D}{x^2+4}

(edited 9 years ago)

Reply 4

JordanSHowarth

I'm really struggling here. Is there anybody who can give me a walk through for the two questions?

Reply 5

davros

Original post by JordanSHowarth

The two answers that I have arrived at are:

1) 6x / x^4-16 = (3/2)/(x^2-4) +(3/2)/(x^2+4) - doesn't seem right to me, with a fraction over the fraction?

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

1) Can't work because if you add those 2 fractions together you get a 6x^2 on top, not 6x! brianeverit has shown you the starting point for this.

2) Again, try adding those 2 fractions together - it doesn't work, does it? Why doesn't it work - because x^3 has a higher degree than the denominator! So can you use poiynomial division first to get something like Ax + B + (another bit that does need partial fractions)?

Reply 6

brianeverit

Original post by JordanSHowarth

I'm really struggling here. Is there anybody who can give me a walk through for the two questions?

Does this help?

Partial Fractions.pdf41.8KB

Reply 7

JordanSHowarth

Thanks Brian.

The first is a quadratic containing partial fraction

The second linear that requires polynomial division because of a higher numerator right?

I will give this a go a little later. I believe I have the correct answers. I just need to be able to show how I arrived there. Thanks for your help :smile: