6 ms·

1) 3x-1/[(x+2)(1-x+x^2)] 2) 1/(x^3+1) 3) 2x^2-14x+8/[(x^2+3x-2)(x-3)] In 1 and 3 are there supposed to be extra brackets? It's really hard to tell exact

1) 3x-1/[(x+2)(1-x+x^2)] 2) 1/(x^3+1) 3) 2x^2-14x+8/[(x^2+3x-2)(x-3)] In 1 and 3 are there supposed to be extra brackets? It's really hard to tell exactly what these are given that you've used no formatting. Secondly, why are you trying to solve these? Where did they come from? Solving the first one, assuming the numerator is supposed to be (3x-1), then we proceed as follows. When we add fractions like a/b + c/d the result is (ad+bc)/bd. In these questions we assume we have the result of such an addition and try to compute all the terms. Here we have a denominator of (x+2)(1-x+x^2) so we assume b=x+2 and d=1-x+x^2. that means we have the numerator formula is: (3x-1) = a(1-x+x^2) + c(x+2) In this case a and c are potentially polynomials (in fact by looking at the degrees c must be a polynomial of degree 1 larger than a). Now just try setting a and c to small polynomials, c being 1 degree larger, and compute what you get on the right hand side. Does it look like the left hand side? Probably not. What do you need to change? How can you make it better. Jiggle it around a bit and see what happens. Tell me, what if a=(x+1) and c=(x^2-2) : what does the RHS evaluate to? Why is it wrong? How can you change a and c to make it better. With regards the second question, are you aware that x^3+1 factorizes? Find a value of x that makes x^3+1 equal zero. What does that tell you about factors of x^3+1? That's enough for now. I've give you a lot of questions. Usually you don't answer them all, just ignoring all but one. I'll see what you do this time. Also, I have to go now - I'll be back tomorrow.

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.
      - 16y ago

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.
    - 16y ago