4 ms·

in 1 and 3 those extra square brackets aren't necessary. secondly i'am trying to solve this as it will help in working out laplace transforms in the future. i

by pencil 16y ago

in 1 and 3 those extra square brackets aren't necessary. secondly i'am trying to solve this as it will help in working out laplace transforms in the future. i went through the partial fraction tutorial in wikipedia and khan acadamy and i'am able to solve the following problems. 1) x/(x+1)(x-4) 2)3x+1/x^2-6x+8 it appears that the above problems are straight forward.all i have to do is solve for 'A' and 'B' after writing them as A/x+1 + B/x-4. but the problems which i have posted are not .i've tried to jiggle it around a lot of times and going no where. with regards to the second question i really don't know how on earth to factorize x^3+1 that's in the denominator. lastly i'll be really happy if you wanna test my ability in math.i'll answer your questions if i'am capable of so that you can suggest what needs to be done next. In fact i believe in learning math the hard way!!!

RiderOfGiraffes 16y ago
You've got real problems with brackets. Major problems that are going to cause significant difficulties down the line. I'll bet the square brackets are necessary, and you need more brackets for the numerator. Your number 1 above you quote as: x/(x+1)(x-4) I'll bet you don't mean that. When you have multiplication and division at the same level, you evaluate left-to-right. The above is the same as: [ x / (x+1) ] (x-4) which is the same as: x (x-4) / (x+1) which I'll bet is not what you intended. The second - as you quote it - is: 3x+1/x^2-6x+8 Division binds more closely than addition and subtraction, while addition and substraction are then read from left-to-right. What you quote is therefore the same as: 3x + 1/x^2 - 6x + 8 which is 3x + (1/x^2) - 6x + 8 which simplifies to (1/x^2) - 3x + 8 I'll bet that's not what you meant. Now let's turn to the first of the ones you originally asked about: 3x-1/[(x+2)(1-x+x^2)] I'll bet you mean: (3x-1) / [(x+2)(1-x+x^2)] and I'll bet the square brackets are necessary. And you are right that all you need to do is solve for A and B after writing it as A/(x+2) + B/(1-x+x^2). So remembering that a/b + c/d = (ad+bc)/bd, what does A/(x+2) + B/(1-x+x^2) equal?
- pencil 16y ago
  all right the brackets concept makes sense. remembering a/b + c/d = (ad+bc)/bd A/(x+2) + B/(1-x+x^2) = A(1-x+x^2) + B(x+2)/(x+2)(1-x+x^2) correct??
  - RiderOfGiraffes 16y ago
    Yes, except you forgot the brackets again. You should have: [ A(1-x+x^2) + B(x+2) ] / [ (x+2)(1-x+x^2) ] You probably knew that, but this is really, really important. Not getting it right now will confuse you a great deal later. However, now you have this as your numerator: A(1-x+x^2) + B(x+2) [Eqn *] Look again at your original problem - what do you want the numerator to be? - Expand out Eqn * to remove the brackets. - What values can you assign to A and B to make Eqn * be the numerator you want? As a hint, A and B are not simply numbers.
    - pencil 16y ago
      first of all i don't know the significance of replacing the numerator with capital letters when decomposing partial fractions.i'am not aware of the practical implications of these problems.all i know is it'll come in handy when solving laplace transforms in the future which is used in physics/electrical etc.(that's what i'am after). and if you are curious to know why i wanna learn these..well i don't have a proper reason.i simply want to!! i looked back at the original problem but i'am unable to figure out the values to assign to A and B.
      - RiderOfGiraffes 16y ago
        OK, to recap. You want to decompose this: [ 3x-1 ] / [ (x+2)(1-x+x^2) ] into fractions. In other words, you want to find A and B such that: A/(x+2) + B/(1-x+x^2) = [ 3x-1 ] / [ (x+2)(1-x+x^2) ] Apart from getting the brackets wrong, you said (correctly) that the left hand side is equal to this: [ A(1-x+x^2) + B(x+2) ] / [ (x+2)(1-x+x^2) ] So now you need to solve: [ A(1-x+x^2)+B(x+2) ] / [ (x+2)(1-x+x^2) ] = (3x-1) / [ (x+2)(1-x+x^2) ] The denominators are equal, so you just need to make the numerators equal. So tell me - what equation do you have to solve?

by pencil 16y ago

RiderOfGiraffes 16y ago

You've got real problems with brackets. Major problems that are going to cause significant difficulties down the line. I'll bet the square brackets are necessary, and you need more brackets for the numerator. Your number 1 above you quote as: x/(x+1)(x-4) I'll bet you don't mean that. When you have multiplication and division at the same level, you evaluate left-to-right. The above is the same as: [ x / (x+1) ] (x-4) which is the same as: x (x-4) / (x+1) which I'll bet is not what you intended. The second - as you quote it - is: 3x+1/x^2-6x+8 Division binds more closely than addition and subtraction, while addition and substraction are then read from left-to-right. What you quote is therefore the same as: 3x + 1/x^2 - 6x + 8 which is 3x + (1/x^2) - 6x + 8 which simplifies to (1/x^2) - 3x + 8 I'll bet that's not what you meant. Now let's turn to the first of the ones you originally asked about: 3x-1/[(x+2)(1-x+x^2)] I'll bet you mean: (3x-1) / [(x+2)(1-x+x^2)] and I'll bet the square brackets are necessary. And you are right that all you need to do is solve for A and B after writing it as A/(x+2) + B/(1-x+x^2). So remembering that a/b + c/d = (ad+bc)/bd, what does A/(x+2) + B/(1-x+x^2) equal?

pencil 16y ago
all right the brackets concept makes sense. remembering a/b + c/d = (ad+bc)/bd A/(x+2) + B/(1-x+x^2) = A(1-x+x^2) + B(x+2)/(x+2)(1-x+x^2) correct??
- RiderOfGiraffes 16y ago
  Yes, except you forgot the brackets again. You should have: [ A(1-x+x^2) + B(x+2) ] / [ (x+2)(1-x+x^2) ] You probably knew that, but this is really, really important. Not getting it right now will confuse you a great deal later. However, now you have this as your numerator: A(1-x+x^2) + B(x+2) [Eqn *] Look again at your original problem - what do you want the numerator to be? - Expand out Eqn * to remove the brackets. - What values can you assign to A and B to make Eqn * be the numerator you want? As a hint, A and B are not simply numbers.
  - pencil 16y ago
    first of all i don't know the significance of replacing the numerator with capital letters when decomposing partial fractions.i'am not aware of the practical implications of these problems.all i know is it'll come in handy when solving laplace transforms in the future which is used in physics/electrical etc.(that's what i'am after). and if you are curious to know why i wanna learn these..well i don't have a proper reason.i simply want to!! i looked back at the original problem but i'am unable to figure out the values to assign to A and B.
    - RiderOfGiraffes 16y ago
      OK, to recap. You want to decompose this: [ 3x-1 ] / [ (x+2)(1-x+x^2) ] into fractions. In other words, you want to find A and B such that: A/(x+2) + B/(1-x+x^2) = [ 3x-1 ] / [ (x+2)(1-x+x^2) ] Apart from getting the brackets wrong, you said (correctly) that the left hand side is equal to this: [ A(1-x+x^2) + B(x+2) ] / [ (x+2)(1-x+x^2) ] So now you need to solve: [ A(1-x+x^2)+B(x+2) ] / [ (x+2)(1-x+x^2) ] = (3x-1) / [ (x+2)(1-x+x^2) ] The denominators are equal, so you just need to make the numerators equal. So tell me - what equation do you have to solve?