3 ms·
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 necess
by 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 agoall 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 agoYes, 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 agofirst 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 agoOK, 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?
- pencil 16y agoi get A(1-x+x^2) + B(x+2) = 3x-1
- RiderOfGiraffes 16y agoExcellent. + Expand B(x+2) + Expand A(1-x+x^2) What do you get? You need to play with these sorts of equations and see what happens - see what you get.
- pencil 16y agoexpand B(x+2)?? but how??