Polynomial division

  • Thread starter ssb
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  • #1
ssb
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I just cant remember how to do this!!! Ive been to several sites suggesting synthetic division and other guides on polynomial division but i cant get it into my head and its driving me wild. :surprised

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

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

Heres my work ive 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 im stuck. I went further and solved A = 16 and B = -8 and pluged it in to get

16/(x+2) - 8/(x+1) but this doesnt 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
 
Last edited:

Answers and Replies

  • #2
morphism
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Try solving for A and B again! I thin you made a mistake somewhere. :smile:
 
  • #3
Hurkyl
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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

[tex]
\frac{8x - 8}{x^2 + 3x + 2}
[/tex]

in the form

[tex]\frac{A}{x+2} + \frac{B}{x+1}.[/tex]

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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  • #4
ssb
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my goal was to simplify it enough so i can integrate it easially. im going to work on it some more right now and ill report back.
 
  • #5
ssb
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Ok heres 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 doesnt 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 im left with nothing!

Any insite this time? Thanks so much so far
 
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  • #6
ssb
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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
 
  • #7
ssb
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Nope doesnt work. im stuck again :-( I ended getting A = 0 and B = 4. A = 0 doesnt really make a whole lotta sense.
 
  • #8
ssb
120
0
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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