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You are not logged in. #1 20131120 07:20:58
repeated linear factors, (i think)hello im having a problem with the last part of this assignment that I have. #2 20131120 08:26:31
Re: repeated linear factors, (i think)hi Simon, Your first step is ok So now compare the x^2 terms, the x terms and the nonx terms x^2 : AB = 4 .......................1 x : 4A + B + C = 14 .............2 nonx: 4A + 2B + C = 0 ...............3 With three unknowns and three equations this should be possible. Look for an easy way to find a letter. Thinks If I use 2 and 3 to eliminate C then I've got a pair of equations with just A and B. So subtract 2 from 3 8A + B = 14 ..................4 You have to watch the signs here! Now if I add equation 1 to 4 the Bs will go: 9A = 18 => A = 2 Now I can get B from equation 1 2  B = 4 => B = 2 Now I can get C from equation 3 4x2 + 2 x 2 + C = 0 => C = 4 And there you have it. Bob ps. Alternative method that works for this example but may not in general. If you put x = 2 the (2x) factors contribute zero so we get 14 x 2  4 x 2 x 2 = C x 3 => 28  16 = 3C => 12 = 3C => C = 4 Put x = 1 and the (1+x) factors become zero 14 x 1  4 x 1 x 1 = A(21)^2 => 14 4 = 9A => 18 = 9A => A = 2 There's no easy value to get B so choose the easiest you can. Put x = 0 and use the values we know for A and C 0 = 2 x 2^2 + B x 1 x 2 + 4 x 1 => 0 = 8 + 2B + 4 => 0 = 4 + 2B => B = 2 Last edited by bob bundy (20131120 08:40:30) You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #3 20131120 08:51:04
Re: repeated linear factors, (i think)what I have to show is A+B or AB, I thought it was AB as I got a B from the repeated factors bit. #4 20131120 09:47:46
Re: repeated linear factors, (i think)hi Simon, You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #5 20131120 10:47:33
Re: repeated linear factors, (i think)hello bob, #6 20131120 20:36:38
Re: repeated linear factors, (i think)hi Simon,
Or you can use the x^2 terms:
Put in the value of A You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #7 20131121 00:31:25
Re: repeated linear factors, (i think)hello bob #10 20131121 06:15:41
Re: repeated linear factors, (i think)So what was the algebraic fraction at the start of the question? If this is correct, then stop worrying; you have correctly identified the three partial fractions: and you have the repeated factor ok. You can get A and C quickly by setting x = 1 and x = 2 You have the correct answers. There is no trick for getting B like this. I know of two ways to get B and I've shown you both. Choose the one you like best and finish the exercise. Bob Last edited by bob bundy (20131121 06:51:38) You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei 