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sat_chen
 one year ago
Best ResponseYou've already chosen the best response.0hold on i didnt know you couldnt post pic from iphone jsut 1 sec

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1draw your question, maybe...

sat_chen
 one year ago
Best ResponseYou've already chosen the best response.0i have done a lot of the steps so just 1 sec

sat_chen
 one year ago
Best ResponseYou've already chosen the best response.0but heres the question dw:1359962006954:dw

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1that will be pretty exhausting..

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1x^4 (x+3) A/x+B/x^2+C/x^3+D/x^4 +E/(x+3)

sat_chen
 one year ago
Best ResponseYou've already chosen the best response.0do you see my steps ill try posting another pic

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1yes, D=1, E=4 are correct.

sat_chen
 one year ago
Best ResponseYou've already chosen the best response.0yea i got that far phew but how do i solve for the A B and C after that?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1you'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...

sat_chen
 one year ago
Best ResponseYou've already chosen the best response.0can you show me what you mean i dont get what you mean by that

sat_chen
 one year ago
Best ResponseYou've already chosen the best response.0should really have listened in my algebra classes sorry :(

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1\(Ax^3(x+3)+Bx^2(x+3)+Cx^3(x+3)+(x+3)+4x^4\\ =9x^4+17x^3+3x^28x+3\) put x = 1, to get 1 equation in A,B, C

sat_chen
 one year ago
Best ResponseYou've already chosen the best response.0it should be Cx (x+3)

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1another method is comparing the coefficients to get 3 equations... yes, typo...it should be Cx(x+3)

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1like coefficient of x^4 will be A+4 and no right side, its 9 so, A+4= 9 gives A=5

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1i think you should proceed with comparing the coefficients..

sat_chen
 one year ago
Best ResponseYou've already chosen the best response.0so if you plug in x = 1 you get 4A + 4B+ 4C = 16

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1i think comapring the coefficients will give you answer much faster, if you have understood it.

sat_chen
 one year ago
Best ResponseYou've already chosen the best response.0sorry internet disconnected agian

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1coefficient of x^3 = 3A+B = 15+B on left on right its 17 so, 15+B = 17 gives B=3

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1forget about substitution...

sat_chen
 one year ago
Best ResponseYou've already chosen the best response.0ok 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

sat_chen
 one year ago
Best ResponseYou've already chosen the best response.0where did you get thedw:1359963485018:dw

sat_chen
 one year ago
Best ResponseYou've already chosen the best response.0then i also guess i dont get how you got the A + 4 either because my logic was wrong

sat_chen
 one year ago
Best ResponseYou've already chosen the best response.0not getting the dw:1359963731639:dw

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1\(Ax^3(x+3)+Bx^2(x+3)+Cx(x+3)+(x+3)+4x^4\\ =9x^4+17x^3+3x^28x+3 \\ ~ \\Ax^4+3Ax^3 +Bx^3+3Bx^2+Cx^2+3Cx+x+3+4x^4= \\ =9x^4+17x^3+3x^28x+3\)

sat_chen
 one year ago
Best ResponseYou've already chosen the best response.0do you have a formula or something that tells us that or how does that work?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1\(Ax^3(x+3)+Bx^2(x+3)+Cx(x+3)+(x+3)+4x^4\\ =9x^4+17x^3+3x^28x+3 \\ ~ \\Ax^4+3Ax^3 +Bx^3+3Bx^2+Cx^2+3Cx+x+3+4x^4= \\ =9x^4+17x^3+3x^28x+3 \\ ~ \\ (A+4)x^4+(3A+B)x^3+....=...\)

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1comapring the cefficiens on left and right, A+4=9 3A+B=17 did you get this ?

sat_chen
 one year ago
Best ResponseYou've already chosen the best response.0its not showing it in nice formatting so im writing it down just 1 sec

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1Ax^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
 one year ago
Best ResponseYou've already chosen the best response.1i just combined like terms on left..
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