3
$\begingroup$

I've been starting to get the hang on partial fractions, whilst I've been able to do most of the basic ones, this kept causing some issues so I assumed:

  1. I'm using the wrong method
  2. I'm converting values when I shouldn't be

Before going on I'll post the question:

(a) Express $\frac{x^2 + 6x + 7}{(x - 3) (x^2 +2x + 2))}$ in partial fractions

After seeing that this question only held 3 marks out of 100 I thought that it was relatively simple. However the values that I was receiving were nothing like what I would normally get. I could post my results but honestly I wrote so many it's irrelevant.

What I initially did was turn the $(x^2 +2x +2)$ into $(x+2)(x+1)$ which would leave us with: $$\frac{x^2 + 6x + 7}{(x - 3) (x+2)(x+1)}$$

From there I used the usual method by placing the equation like so: $$\frac{A}{(x-3)}+\frac{B}{(x+2)}+\frac{C}{(x+1)}$$

After that I found the LCM and starts cutting off terms by replacing $x$ with a specific value: $$A(x+2)(x+1) + B(x-3)(x+1) + C(x-3)(x+2)$$

The rest is basically history, I can barely understand what I was even trying to do. Feel free to guide me to the right direction

Mistakes pointed out:

  1. My factorisation is incorrent for $$x^2 + 2x +2 = (x + 2)(x+1)$$

Extra Attempts

My second attempt was done using the method for quadratic factors inside the denominator. So it's now: $$\frac{A}{(x-3)}+\frac{Bx+C}{(x^2 + 2x + 2)}$$

From here on I think I'm meant to find the LCM by doing the following: $$(Bx+C)(x-3) + A(x^2 + 2x +2)$$

I then substitute $x$ with 3 in order to find the value of $A$ which would end up like so: $$16 + 6x = 17A$$

After that I'm not entirely sure if it's correct(highly doubt so)

However not much has changed in terms of getting a viable answer

$\endgroup$
14
  • 1
    $\begingroup$ $(x+2)(x+1)=x^2+3x+2$ so your initial factorisation is incorrect $\endgroup$
    – Mufasa
    Commented Sep 7, 2014 at 12:13
  • $\begingroup$ @Mufasa Woops you are right, my bad! That's what I get for not actually testing it beforehand. I've went ahead and pointed out the mistake in the original post. I'll give it another go and see what I can come up with. $\endgroup$
    – Juxhin
    Commented Sep 7, 2014 at 12:15
  • 1
    $\begingroup$ I have posted them in the question Mufasa it's in the beginning. However Mary Star pointed out a critical mistake and provided me with a good template to work on. So I'll start off by working it out using her answer $\endgroup$
    – Juxhin
    Commented Sep 7, 2014 at 12:28
  • 1
    $\begingroup$ When you put in $x=3$, you can't get an equation that still has $x$ in it. You should be able to get the value of $A$. Then put in some other simple values for $x$, say, $x=0$ and $x=1$, to get two equations for $B$ and $C$. $\endgroup$
    – Gerry Myerson
    Commented Sep 7, 2014 at 12:44
  • 1
    $\begingroup$ When you used $x=3$ you forgot to replace the $x$ in $16+6x=17A$ which would have led you to $16+18=17A\implies 34=17A$ $\endgroup$
    – Mufasa
    Commented Sep 7, 2014 at 12:45

1 Answer 1

Reset to default
2
$\begingroup$

$$x^2+2x+2=0 \Rightarrow \Delta=4-4 \cdot 2=4-8=-4<0$$

So it has no real roots.

Therefore, to express $\frac{x^2 + 6x + 7}{(x - 3) (x^2 +2x + 2)}$ in partial fractions we do the following:

$$\frac{x^2 + 6x + 7}{(x - 3) (x^2 +2x + 2)}=\frac{A}{x-3}+\frac{Bx+c}{x^2+2x+2}$$

(The polynomial at the numerator has to be one degree smaller than the degree of the polynomial of the denominator.)

EDIT:

$$\frac{x^2 + 6x + 7}{(x - 3) (x^2 +2x + 2)}=\frac{A}{x-3}+\frac{Bx+c}{x^2+2x+2} \\ \Rightarrow \frac{x^2 + 6x + 7}{(x - 3) (x^2 +2x + 2)}=\frac{A(x^2+2x+2)+(Bx+C)(x-3)}{(x-3)(x^2+2x+2)} \\ \Rightarrow x^2+6x+7=Ax^2+2Ax+2A+Bx^2-3Bx+Cx-3C \\ \Rightarrow x^2+6x+7=(A+B)x^2+(2A-3B+C)x+(2A-3C) $$ Now you have to solve the following system:

$$A+B=1 \\ 2A-3B+C=6 \\ 2A-3C=7$$

$\endgroup$
4
  • 1
    $\begingroup$ Oh alright, I had the same idea but I executed it in the wrong many. This is very helpful, I should be able to work it out like this. Thanks Mary $\endgroup$
    – Juxhin
    Commented Sep 7, 2014 at 12:27
  • 1
    $\begingroup$ If I may ask you to give my original post another look under Extra Attempts to see what I must have done. Sorry! $\endgroup$
    – Juxhin
    Commented Sep 7, 2014 at 12:38
  • $\begingroup$ I added some more information at my answer!! $\endgroup$
    – Mary Star
    Commented Sep 7, 2014 at 12:42
  • 2
    $\begingroup$ Oh wow thanks alot! This makes alot more sense now $\endgroup$
    – Juxhin
    Commented Sep 7, 2014 at 12:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .