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