Polynomial Division: Solving Denominator > Numerator

AI Thread Summary
The discussion centers on the challenges of performing polynomial division when the denominator is larger than the numerator. The original poster struggles with using the partial fractions method to express (8x-8)/(x^2+3x+2) correctly, initially making a mistake in calculating coefficients A and B. After receiving guidance, they correct their calculations and successfully determine A = -16 and B = 24. The conversation then shifts to another problem involving a perfect square in the denominator, where the poster encounters difficulties in solving for A and B but eventually resolves the issue. The thread highlights the importance of careful algebraic manipulation in polynomial division and partial fractions.
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I just can't remember how to do this! I've been to several sites suggesting synthetic division and other guides on polynomial division but i can't get it into my head and its driving me wild.

(8x-8)/(x^2+3x+2)

synthetic division doesn't work here because the denominator is larger than the numerator... right?

Heres my work I've done w/ that P/Q method:

(x^2+3x+2) = (x+2)(x+1)

therefore use the format:

8x-8 = A/(x+2) + B(x+1)
and solve for A and B

This is where I am stuck. I went further and solved A = 16 and B = -8 and pluged it into get

16/(x+2) - 8/(x+1) but this doesn't equal (8x-8) / (x^2+3x+2) !

Im stuck! can someone please enlighten me? Thanks a ton!

OR: If anyone has a good helpful website that they know of that shows polynomial division when the Denominator is larger than the numerator id appreciate it. I can only find the sites where the numerator is larger than the denominator
 
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Try solving for A and B again! I thin you made a mistake somewhere. :smile:
 
What I think you're trying to do is called partial fractions. Polynomial division in this case is very easy: the quotient is 0 and the remainder is 8x-8. :wink:


Your goal, presumably, is to write the quotient

<br /> \frac{8x - 8}{x^2 + 3x + 2}<br />

in the form

\frac{A}{x+2} + \frac{B}{x+1}.

So you want those to be equal, right? ...

(clearly 8x-8 = A/(x+2) + B(x+1) was a typo -- what did you really mean by that?)
 
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my goal was to simplify it enough so i can integrate it easially. I am going to work on it some more right now and ill report back.
 
Ok here's my new work:

8x-8 = A(x+2) + B(x+1)

Sub in -2 for x to get B
8(-2)-8 = A(-2+2) + B(-2+1)

Simplify to
B = 24

then i subed in -1 for x to solve for A
8(-1)-8 = A(-1+2) + B(-1+1)

simplify to
A = 16 <------------ Thats where my mistake was! its -16 not 16!

OMG YAY YAY YAY YAY YAY I CAN DO MATH!

THANKYOU ALL SO MUCH

But this brings up another question. I tried to do a problem with a perfect square in the denominator and the method doesn't work... :-(

This is the one i made right now

(8x+12) / (x+1)^2

so:
[(x+1) (x+1)^2 (8x+12)] / (x+1)^2 = [A/(x+1) + B/ (x+1)^2 (x+1)(x+1)^2]

simplifies to

(x+1) (8x+12) = A(x+1)^2 +B(x+1)

when i try to solve for A and B, no matter what number i sub in for X they both end up zeroing out and I am left with nothing!

Any insite this time? Thanks so much so far
 
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Ok in my attempt i multiplied both sides by both terms. I tried it by multiplying both sides by one term at a time and i think its working... ill report back
 
Nope doesn't work. I am stuck again :-( I ended getting A = 0 and B = 4. A = 0 doesn't really make a whole lotta sense.
 
Actually I just figured it out. I learned that my alegebra needs some work.

A = 8
B = 4
THANKS ill be back later with more questions!
 
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