Specifically equating coefficients with like terms after "foil"ing, I see this in other problems and I'm not sure why its there or why it makes sense exactly.
Also, I am comfortable with imaginaries, but doesnt that needlessly make it more complicated? or from your perspective do you find it simpler by combining the i's, but what happens whenyou go back to reals when not all i's become an even power of i?
Using x=i gives you $\displaystyle 1=4(Ci+D)$.
On the left you have $\displaystyle 1+0i$
On the right you have $\displaystyle 4D+4Ci$
Equating real and imaginary parts you see that 4D=1 and 4C=0.
I may get back to your other points later..
Starting with:
$\displaystyle 1=(Ax+B)(x^2+1)+(Cx+D)(x^2+5)$
multiply and add the polynomials to get:
$\displaystyle 1=(Ax^3+Bx^2+Ax+B)+(Cx^3+Dx^2+5Cx+5D)$
$\displaystyle 1=(A+C)x^3+(B+D)x^2+(A+5C)x+(B+5D)$
You need these to be equal as polynomials, so in other words, all the coefficients must be equal:
$\displaystyle 0=A+C$
$\displaystyle 0=B+D$
$\displaystyle 0=A+5C$
$\displaystyle 1=B+5D$
Usually when you see equations like this, you're trying to figure out which values of x make the equation true, but here you want 1, which is $\displaystyle 0x^3+0x^2+0x+1$, to be the same polynomial as $\displaystyle (A+C)x^3+(B+D)x^2+(A+5C)x+(B+5D)$, so that they are equal for all values of x.
The rest is using these four equations and the two equations from the previous part to solve for A, B, C, and D.
Step 4 is actually unnecessary, by the way; the four equations in step 5 are all you need to solve for the four unknowns A, B, C, and D.
- Hollywood