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• 1 Post By hollywood

# Thread:

1. ## Partial fractions

Please explain the step in the green box(#5), I remember doing this in algebra 2 but not this specifically.
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2. ## Re: Partial fractions

Which bit don't you understand?

Incidentally substituting x=i (as well as x=0 and x=1) is quite handy too if you're happy with complex numbers. It gives you C and D and ten using the firs two equations you can also get A and B.

3. ## Re: Partial fractions

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?

4. ## Re: Partial fractions

Using x=i gives you $1=4(Ci+D)$.

On the left you have $1+0i$

On the right you have $4D+4Ci$

Equating real and imaginary parts you see that 4D=1 and 4C=0.

I may get back to your other points later..

5. ## Re: Partial fractions

Starting with:

$1=(Ax+B)(x^2+1)+(Cx+D)(x^2+5)$

multiply and add the polynomials to get:

$1=(Ax^3+Bx^2+Ax+B)+(Cx^3+Dx^2+5Cx+5D)$
$1=(A+C)x^3+(B+D)x^2+(A+5C)x+(B+5D)$

You need these to be equal as polynomials, so in other words, all the coefficients must be equal:

$0=A+C$
$0=B+D$
$0=A+5C$
$1=B+5D$

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 $0x^3+0x^2+0x+1$, to be the same polynomial as $(A+C)x^3+(B+D)x^2+(A+5C)x+(B+5D)$, 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

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