anonymous
  • anonymous
Partial integration question
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
hold on i didnt know you couldnt post pic from iphone jsut 1 sec
hartnn
  • hartnn
draw your question, maybe...
anonymous
  • anonymous
i have done a lot of the steps so just 1 sec

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anonymous
  • anonymous
but heres the question |dw:1359962006954:dw|
hartnn
  • hartnn
that will be pretty exhausting..
hartnn
  • hartnn
x^4 (x+3) A/x+B/x^2+C/x^3+D/x^4 +E/(x+3)
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
do you see my steps ill try posting another pic
hartnn
  • hartnn
yes, D=1, E=4 are correct.
anonymous
  • anonymous
yea i got that far phew but how do i solve for the A B and C after that?
hartnn
  • hartnn
you'll need to put 3 different values of x say x=1,-1,2 then you get 3 equations in A,B, C with 3 unknowns...
anonymous
  • anonymous
can you show me what you mean i dont get what you mean by that
anonymous
  • anonymous
should really have listened in my algebra classes sorry :(
hartnn
  • hartnn
\(Ax^3(x+3)+Bx^2(x+3)+Cx^3(x+3)+(x+3)+4x^4\\ =9x^4+17x^3+3x^2-8x+3\) put x = 1, to get 1 equation in A,B, C
anonymous
  • anonymous
it should be Cx (x+3)
hartnn
  • hartnn
another method is comparing the co-efficients to get 3 equations... yes, typo...it should be Cx(x+3)
hartnn
  • hartnn
like co-efficient of x^4 will be A+4 and no right side, its 9 so, A+4= 9 gives A=5
hartnn
  • hartnn
i think you should proceed with comparing the co-efficients..
anonymous
  • anonymous
so if you plug in x = 1 you get 4A + 4B+ 4C = 16
hartnn
  • hartnn
i think comapring the co-efficients will give you answer much faster, if you have understood it.
anonymous
  • anonymous
sorry internet disconnected agian
hartnn
  • hartnn
co-efficient of x^3 = 3A+B = 15+B on left on right its 17 so, 15+B = 17 gives B=3
hartnn
  • hartnn
forget about substitution...
anonymous
  • anonymous
ok so what you are saying is that for A you A + 4 = 9 then for B would you B + 3 =17 then for C would you C + 2 =3
anonymous
  • anonymous
where did you get the|dw:1359963485018:dw|
hartnn
  • hartnn
it should be B+3A =17
anonymous
  • anonymous
then i also guess i dont get how you got the A + 4 either because my logic was wrong
hartnn
  • hartnn
|dw:1359963578884:dw|
anonymous
  • anonymous
not getting the |dw:1359963731639:dw|
hartnn
  • hartnn
\(Ax^3(x+3)+Bx^2(x+3)+Cx(x+3)+(x+3)+4x^4\\ =9x^4+17x^3+3x^2-8x+3 \\ ~ \\Ax^4+3Ax^3 +Bx^3+3Bx^2+Cx^2+3Cx+x+3+4x^4= \\ =9x^4+17x^3+3x^2-8x+3\)
anonymous
  • anonymous
do you have a formula or something that tells us that or how does that work?
hartnn
  • hartnn
\(Ax^3(x+3)+Bx^2(x+3)+Cx(x+3)+(x+3)+4x^4\\ =9x^4+17x^3+3x^2-8x+3 \\ ~ \\Ax^4+3Ax^3 +Bx^3+3Bx^2+Cx^2+3Cx+x+3+4x^4= \\ =9x^4+17x^3+3x^2-8x+3 \\ ~ \\ (A+4)x^4+(3A+B)x^3+....=...\)
hartnn
  • hartnn
comapring the c-efficiens on left and right, A+4=9 3A+B=17 did you get this ?
anonymous
  • anonymous
its not showing it in nice formatting so im writing it down just 1 sec
hartnn
  • hartnn
Ax^3(x+3)+Bx^2(x+3)+Cx(x+3)+(x+3)+4x^4=9x^4+17x^3+3x^2−8x+3 Ax^4+3Ax^3+Bx^3+3Bx^2+Cx^2+3Cx+x+3+4x^4=9x^4+17x^3+3x^2−8x+3 (A+4)x^4+(3A+B)x^3+....=9x^4+17x^3+...
hartnn
  • hartnn
i just combined like terms on left..

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