<s>[INST] Basic partial fractions issue 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:


*

*I'm using the wrong method

*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:


*

*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
 [/INST] $$x^2+2x+2=0 \Rightarrow \Delta=4-4 \cdot 2=4-8=-4<0$$
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:
$$A+B=1 \\ 2A-3B+C=6 \\ 2A-3C=7$$
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