Partial fractions question help pls asap

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Htx_x346
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#1
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#1
How do you work out ax +b?
Is the fastest method long division cause that’ll take me like 5 minsName:  4D9F070D-0DA7-4D15-9E95-EAEFE5347BB7.jpeg
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Last edited by Htx_x346; 1 month ago
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orangejuze
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#2
long division is fun 😤
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Htx_x346
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(Original post by orangejuze)
long division is fun 😤
You like orange juice so your opinion doesn't count 🙅*♀️
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orangejuze
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(Original post by Htx_x346)
You like orange juice so your opinion doesn't count 🙅*♀️
dont tell me you an apple juice lover🙁
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Htx_x346
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(Original post by orangejuze)
dont tell me you an apple juice lover🙁
Yeah apple juice>>🤩🤩

Please get off my thread- I don't want to be associated with orange juice lovers😒😤😤
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Htx_x346
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(Original post by orangejuze)
dont tell me you an apple juice lover🙁
Yk I was joking btw😭
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mqb2766
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#7
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(ax + b)*(x^2 + 2x - 8) + r(x) = 5x^3 + 8x^2 + ...
fairly trivially matching the cubic and quadratic coeffs
a = 5
b = -2
where r(x) is the linear remainder.
Last edited by mqb2766; 1 month ago
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orangejuze
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(Original post by Htx_x346)
Yk I was joking btw😭
i know, i was busy drinking orange juice 🥵
Last edited by orangejuze; 1 month ago
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jon nicholls
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Can I just point out that D should be over x+4.
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jon nicholls
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So by inspection A=5 as you just need to divide the dominant term in the numerator by the dominant term in the denominator 5x^3 / x^2.

B is much tougher to find. I've never been brilliant with long division so I prefer .....

 



\dfrac{5x^3+8x^2-46x+26}{x^2+2x-8}\\



  = \dfrac{5x^3+10x^2-40x}{x^2+2x-8}+\dfrac{-2x^2-6x+26}{x^2+2x-8}\\



  = 5x +\dfrac{-2x^2-6x+26}{x^2+2x-8}\\



  = 5x +\dfrac{-2x^2-4x+16}{x^2+2x-8}+\dfrac{-2x+10}{x^2+2x-8}\\



  =5x-2 + \dfrac{10-2x}{(x-2)(x+4)}\\



  =5x-2+\dfrac{1}{x-2}-\dfrac{3}{x+4}


For those who don't like long division Still takes a while.....
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Htx_x346
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(Original post by orangejuze)
i know, i was busy drinking orange juice 🥵



ew.



(Original post by mqb2766)
(ax + b)*(x^2 + 2x - 8) + r(x) = 5x^3 + 8x^2 + ...
fairly trivially matching the cubic and quadratic coeffs
a = 5
b = -2
where r(x) is the linear remainder.
Isnt that basically long division cause you dont know a or b? Sorry this is in italics i cba to change it
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mqb2766
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(Original post by Htx_x346)
Isnt that basically long division cause you dont know a or b? Sorry this is in italics i cba to change it
Of course, but I cant see how it would take 5 minutes to match 2 coefficients like that. a is "write down" and b is simply
8 = 2*a + b
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Htx_x346
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#13
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#13
(Original post by mqb2766)
Of course, but I cant see how it would take 5 minutes to match 2 coefficients like that. a is "write down" and b is simply
8 = 2*a + b
Oh yeah i get you.
once you work out a and b you have to long divide though am i right? To find rx?
especially if its ax^2 or something?
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mqb2766
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(Original post by Htx_x346)
Oh yeah i get you.
once you work out a and b you have to long divide though am i right? To find rx?
especially if its ax^2 or something?
To find the remainder, youd have to subtract (ax+b)(x^2+2x-8) from the original cubic, BUT you only need to do it for the constant and linear term (youve matched the quadratic and cubic terms already). So again, quick.
Last edited by mqb2766; 1 month ago
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Htx_x346
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(Original post by mqb2766)
To find the remainder, youd have to subtract (ax+b)(x^2+2x-8) from the original cubic, BUT you only need to do it for the constant and linear term (youve matched the quadratic and cubic terms already). So again, quick.
Omg yeah ur right, what was i thinking
thank u sm
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