Find the ordered triplet (A,B,C) if: (6x^2 - 8x -20) / [(x+2)(x-3)(x-2)] = (A/(x+2)) + (B/(x-3)) + (C/(x-2)) **hint: mulitply both sides by (x+2)(x-3)(x-2)
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Originally Posted by Mr_Green Find the ordered triplet (A,B,C) if: (6x^2 - 8x -20) / [(x+2)(x-3)(x-2)] = (A/(x+2)) + (B/(x-3)) + (C/(x-2)) **hint: mulitply both sides by (x+2)(x-3)(x-2) Multiply through and you get, Now this is a partial fraction decomposition which can always be achived. Thus this is true for all . If you can solve for If you can solve for If you can solve for
i got A = 1 B = 23.6 C = 6 Could someone check that i got this right?
Nvm, i got the answer and checked it... A = 1 B = 2 C = 3 thanks everyone!
Originally Posted by Mr_Green i got A = 1 B = 23.6 C = 6 Could someone check that i got this right? Hello, Mr Green, TPH had made a small but powerful typo: In If x=3 you can solve for B the exponent of the first 3 should be 2 Then you'll get: B = 2 EB
thanks
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