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

  • 1
    $\begingroup$ $(x+2)(x+1)=x^2+3x+2$ so your initial factorisation is incorrect $\endgroup$ – Mufasa Sep 7 '14 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 Sep 7 '14 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 Sep 7 '14 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 Sep 7 '14 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 Sep 7 '14 at 12:45

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

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.)


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

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

  • 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 Sep 7 '14 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 Sep 7 '14 at 12:38
  • $\begingroup$ I added some more information at my answer!! $\endgroup$ – Mary Star Sep 7 '14 at 12:42
  • 2
    $\begingroup$ Oh wow thanks alot! This makes alot more sense now $\endgroup$ – Juxhin Sep 7 '14 at 12:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.