# Thread: how do you decompose this into partial fraction?

1. ## how do you decompose this into partial fraction?

. I am revising for my end of year exams. Hence, I have been throwing lots of questions lately.

Express $\displaystyle \dfrac{2}{x(x^2+5)}$ in partial fractions.

What are the structures for the corresponding partial fractions in the first place?

Method 1: conventional method of substitution
Method 2: Cover-up method

Which method should I use? How?

PS: my latex skills improved!

2. Originally Posted by stupidguy
. I am revising for my end of year exams. Hence, I have been throwing lots of questions lately.

Express $\displaystyle \frac{2}{x(x^2+5)}$ in partial fractions.

What are the denominators of the partial fractions in the first place?

PS: my latex skills improved!

check those...

YouTube - Partial Fraction Expansion 1

YouTube - Partial Fraction Expansion 2

YouTube - Partial Fraction Expansion 3

3. Originally Posted by stupidguy
Express $\displaystyle \frac{2}{x(x^2+5)}$ in partial fractions.
PS: my latex skills improved!
$$\dfrac{2}{x(x^2+5)}$$ gives $\displaystyle \dfrac{2}{x(x^2+5)}$.
After the update we need to make some changes.
Notice the \dfrac tag for \displaystyle

4. Originally Posted by Plato
$$\dfrac{2}{x(x^2+5)}$$ gives $\displaystyle \dfrac{2}{x(x^2+5)}$.
After the update we need to make some changes.
Notice the \dfrac tag for \displaystyle
Changed!

5. Let $\displaystyle \displaystyle \frac{2}{x(x^2+5)} = \frac{a}{x}+\frac{bx+c}{x^2+5}$.

6. Originally Posted by TheCoffeeMachine
Let $\displaystyle \displaystyle \frac{2}{x(x^2+5)} = \frac{a}{x}+\frac{bx+c}{x^2+5}$.
Is it when I see x the corresponding partial fraction is $\displaystyle \dfrac{a}{x}$?

you treat x as a linear factor?

7. Originally Posted by TheCoffeeMachine
Let $\displaystyle \displaystyle \frac{2}{x(x^2+5)} = \frac{a}{x}+\frac{bx+c}{x^2+5}$.
can I still use cover-up method?

8. Originally Posted by stupidguy
Is it when I see x the corresponding partial fraction is $\displaystyle \frac{a}{x}?$
Yes! The corresponding term for any (unrepeated) linear factor in the denominator is $\displaystyle \frac{a}{x}$.
Is it a rule cos my textbook dun show it?
If $\displaystyle f(x)$ is of less degree than the denominator, then:
The corresponding partial fraction form of $\displaystyle \displaystyle \frac{f(x)}{(x+n_{1})(x+n_{2})(x+n_{3})\cdots(x+n_ {r})}[/Math] is [Math]\displaystyle \frac{a_{1}}{x+n_{1}}+\frac{a_{2}}{x+n_{2}}+\frac{ a_{3}}{x+n_{3}}+\cdots +\frac{a_{r}}{x+n_{r}}$
Originally Posted by stupidguy
can I still use cover-up method?
Of course.

9. a linear factor has the structure of $\displaystyle ax+b$. so you are treating a as 1 and b as 0, right?

If the denominator is 2x, the corresponding partial fraction is $\displaystyle \dfrac{a}{2x}$?

10. did you guys get $\displaystyle \dfrac{2}{5x}-\dfrac{x+1}{5x^2+25}$?

11. Originally Posted by stupidguy
a linear factor has the structure of $\displaystyle ax+b$. so you are treating a as 1 and b as 0, right?
If you have ax+b in the denominator and it's not repeated then you have its corresponding
term is of the form Ax+B, with B = 0, x = 1, yes. So it's just a constant, call it just A.
If the denominator is 2x, the corresponding partial fraction is $\displaystyle \dfrac{a}{2x}$?
That's right.
Originally Posted by stupidguy
did you guys get $\displaystyle \dfrac{2}{5x}-\dfrac{x+1}{5x^2+25}$?
You got $\displaystyle a$ right. I should have cautioned that the cover-up method would only work for finding $\displaystyle a$.

Spoiler:
First, set-up:

$\displaystyle \displaystyle \frac{2}{x(x^2+5)} = \frac{a}{x}+\frac{bx+c}{x^2+5}$

You can see that if you cover-up $\displaystyle (x^2+5)$, then $\displaystyle a = \frac{2}{5}.$ But that is not going to give us b and c, so let's do it in the saver way.

Put the RHS over a common denomintor:

$\displaystyle \displaystyle \frac{2}{x(x^2+5)} = \frac{a(x^2+5)+(bx+c)(x)}{x(x^2+5)}$

We have $\displaystyle 2 = a(x^2+5)+(bx+c)(x)$

[Stop for a moment. Let $\displaystyle x = 0$: Then we have $\displaystyle 2 = 5a$. So $\displaystyle a = \frac{2}{5}$. The cover-up method was really this in disguise. You can see why it doesn't work for finding b and c -- there is no value of x which can make $\displaystyle a(x^2+5)$ equal to zero. So if in general you have a partial fraction form of $\displaystyle \frac{Ax+B}{ax^2+bx+c}$, then you cannot find A and B by covering-up -- in other words, the denominator have to be linear for the cover-up to work.]

We have:

$\displaystyle 2 = ax^2+5a+bx^2+cx$ (expanding the right-hand side).

$\displaystyle \Rightarrow 2 = ax^2+bx^2+cx+5a$ (writing it in terms of descending powers of x).

$\displaystyle \Rightarrow 2 = (a+b)x^2+(c)x+5a$ (factoring out the x's out).

Now, compare the coefficients:

We have $\displaystyle c = 0$.
$\displaystyle 5a = 2[/Math], so [Math]a = \frac{2}{5}$.
$\displaystyle a+b = 0$, and since $\displaystyle a = \frac{2}{5}$, we get $\displaystyle \frac{2}{5}+b = 0$, hence $\displaystyle b= -\frac{2}{5}$.

Therefore $\displaystyle \displaystyle \frac{2}{x(x^2+5)} = \frac{2}{5x}-\frac{2x}{5(x^2+5)}.$

It should of course take a much smaller space to write-up, but that's Mr Beefcake's attempt at being clear.

12. thanks.

BTW, I did not do cover-up for the quadratic factor. I made a careless mistake during substitution, resulting in the wrong answer.

Anyway thanks. Your reply will still be useful for others to learn.