Diatribe....

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lookagain

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Based on this thread:

Because of the 2 exponent in the denominator [the (x+2)2] you will need three terms

\(\displaystyle \dfrac{5x^2+23x+24}{(2x+3)(x+2)^2}=\dfrac{A}{2x+3} + \dfrac{B}{x+2} + \dfrac{Cx+D}{(x+2)^2} \ \ \ \ \ \) <------ That's not the correct form.

You might want to look at
http://www.mathsisfun.com/algebra/partial-fractions.html

This is the correct form:


\(\displaystyle \dfrac{5x^2+23x+24}{(2x+3)(x+2)^2}=\dfrac{A}{2x+3} + \dfrac{B}{x+2} + \dfrac{C}{(x+2)^2} \)


Your link shows an example that is similar enough for that step.
 
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This is the correct form:


\(\displaystyle \dfrac{5x^2+23x+24}{(2x+3)(x+2)^2}=\dfrac{A}{2x+3} + \dfrac{B}{x+2} + \dfrac{C}{(x+2)^2} \)


Your link shows an example that is similar enough for that step.
Thank you for pointing that out. But then I thought that
0 * x + D = D
Are you saying I'm wrong?
 
Thank you for pointing that out. But then I thought that
0 * x + D = D
Are you saying I'm wrong? \(\displaystyle \ \ \ \ \ \ \)Yes! This is fudging.

After you concede that I pointed it out (which is the fact that you used the wrong
form), then do not follow it up with a "but" statement, because that negates you
conceding the point.
 
After you concede that I pointed it out (which is the fact that you used the wrong
form), then do not follow it up with a "but" statement, because that negates you
conceding the point.
Then your saying yes concedes the point that you were saying I was wrong but it was negated by the This is fudging? I'm sorry, I guess I'm confused. Which is it, I was wrong about 0*x+D=D or you are fudging?
 
Then your [sic] saying "yes" concedes the point that you were saying
I was wrong but it was negated by the This is fudging? I'm sorry, I guess I'm confused.

> > > Which is it, I was wrong about 0*x+D=D or you are fudging? < < <

Don't even try to deflect it onto me. Don't continue to try to weasel out of your incompetence.
Your continued cover-up is worse by your initial infraction.

The content in Post # 5 is sufficient. And you know now to exclusively tell students the form that I,
as well as the same content in the link you gave, to use. Drop the matter and don't claim "confusion."
Got it!?
 
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Don't even try to deflect it onto me. Don't continue to try to weasel out of your incompetence.
Your continued cover-up is worse by your initial infraction.

The content in Post # 5 is sufficient. And you know now to exclusively tell students the form that I,
as well as the same content in the link you gave, to use. Drop the matter and don't claim "confusion."
Got it!?
What deflection? You seem to really know what should be done so I'm just asking questions to take advantage of your vast knowledge. If you had rather not share, I think I understand your reticence in speaking to such a (in your eyes) lowly creature as myself. So, taking your suggestion (it was a suggestion wasn't it and not a royal command), I'll just let it go and assume the matter is finished.
 
DISCLAIMER: Beer soaked rambling/opinion/observation/reckoning ahead. Read at your own risk. Not to be taken seriously. In no event shall the wandering math knight-errant Sir jonah in his inebriated state be liable to anyone for special, collateral, incidental, or consequential damages in connection with or arising out of the use of his beer (and tequila) powered views.
Don't even try to deflect it onto me. Don't continue to try to weasel out of your incompetence.
Your continued cover-up is worse by your initial infraction.

The content in Post # 5 is sufficient. And you know now to exclusively tell students the form that I,
as well as the same content in the link you gave, to use. Drop the matter and don't claim "confusion."
Got it!?
Ouch!
Methinks thou art being too harsh Sir lookagain.
You sound like you need to take a chill pill.
A bottle of absinthe or tequila is highly recommended.
You're just going to have to trust me about this one thing.
You need a lot of drinks.
 
This is the correct form:


\(\displaystyle \dfrac{5x^2+23x+24}{(2x+3)(x+2)^2}=\dfrac{A}{2x+3} + \dfrac{B}{x+2} + \dfrac{C}{(x+2)^2} \)
 
This is the correct form:


\(\displaystyle \dfrac{5x^2+23x+24}{(2x+3)(x+2)^2}=\dfrac{A}{2x+3} + \dfrac{B}{x+2} + \dfrac{C}{(x+2)^2} \)
 
I'll just let it go and assume the matter is finished.

Wrong. The matter was finished as soon as I posted this originally:

This is the correct form:


\(\displaystyle \dfrac{5x^2+23x+24}{(2x+3)(x+2)^2}=\dfrac{A}{2x+3} + \dfrac{B}{x+2} + \dfrac{C}{(x+2)^2} \)
 
Wrong. The matter was finished as soon as I posted this originally:

This is the correct form:


\(\displaystyle \dfrac{5x^2+23x+24}{(2x+3)(x+2)^2}=\dfrac{A}{2x+3} + \dfrac{B}{x+2} + \dfrac{C}{(x+2)^2} \)
Well, if it's finished why are you still posting? Now I know there must be some reason know only to super intellectuals such as your self but if you would share the reason with me I would certainly.
 
Well, if it's finished why are you still posting? Now I know there must be some reason know only to super intellectuals
such as your self [sic] but if you would share the reason with me I would certainly.

You were done in post # 7, so mine is the last word. Don't bother posting again. Are you going to continue to act dumb about it,
or are you going to accept your failure and stop posting!?


This is the correct form:


\(\displaystyle \dfrac{5x^2+23x+24}{(2x+3)(x+2)^2}=\dfrac{A}{2x+3} + \dfrac{B}{x+2} + \dfrac{C}{(x+2)^2} \)
 
Lookagain, how do you feel about my correct form of

\(\displaystyle \dfrac{\ \ A}{2x+3} + \dfrac{Bx+C}{\ \ \, \ (x+2)^2}\)?

:cool:

edit: fixed the spacing of the numerator! that was close
 
DISCLAIMER: Beer soaked rambling/opinion/observation/reckoning ahead. Read at your own risk. Not to be taken seriously. In no event shall the wandering math knight-errant Sir jonah in his inebriated state be liable to anyone for special, collateral, incidental, or consequential damages in connection with or arising out of the use of his beer (and tequila) powered views.
Oh I am too drunk for this.
 
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