Question

compare coefficients of:
3x+1=A(x+1)^3 +Bx(x+1)^2 +Cx(x+1) +Dx

Answer 1

Could you put into your own words what you think you're supposed to do in approaching this problem?

Answer 2

write out the coefficients for the values and use simultaneous equations to solve

Answer 3

Looks like you're doing a "partial fraction expansion." Am I right?

Answer 4

correct, i really have no idea how to compare coefficients practically

Answer 5

Looks as tho' we have a lot of multiplication to do here. For example, we have to expand A(x+1)^3 and all the terms that follow this.
(x+1) is a binomial; we can expand it manually, or by using binomial formula, or whatever. Are you familiar with

Answer 6

\[(a+b)^{3}=a ^{3}+3a ^{2}b+3ab ^{2}+b ^{3}\]

Answer 7

??

Answer 8

When i did it i ended up with:
Ax^3+Bx^3 +3Ax^2+2Bx^2+Cx^2 +3Ax+Bx+Dx +C

Answer 9

I'm going to trust your expansion. Take that expansion and group the coefficients of like terms together. For example, up front you have x^3 in two terms; this factors to (x^3)(A+B). Make sense to you?

Answer 10

Please do the same for the rest of your expression. The next power of x is x^2; write out (x^2)( + ) (fill in)

Answer 11

Like (x^3)(A+B+(x^2)(3A+2B+C)+(x)(3A+B+D)+C?

Answer 12

After you've finished doing this, equate this whole expression to\[0x ^{3}+0x ^{2}+3x^1+1\]

Answer 13

Now's the time to compare your coeffs.
On one side you have 0x^3; on the other side, the corresponding x^3 term is (A+B). Do this for each power of x, including x^0.
You'll end up with four simult. equations, which we'll then have to solve for A, B, C and D.

Actually, all you have to do for this particular problem is to obtain those four equations in A, B, C and D; going beyond that would be optional.

Answer 14

BRB (be right back). Please do as much as you can and are willing to do and type your results here.

Answer 15

i'll do it now and post my answers in a few

Answer 16

How's it going ?

Answer 17

not too well, i have c=1, b=3-3A-d and d=\[\frac{ 3A-7 }{ -2 }\] but i'm struggling from there

Answer 18

What I'd write out would be something like this:
A+B=0
3A+2B+C=0

There are 2 more equations to write.

Remember: You don't have to solve this system of equations, but you certainly can if you want. Want to finish finding A, B, C and D?

Answer 19

3=3A+B+D
1=C
i used these to find the values, i'll have to solve to complete the integration

Answer 20

I need to know whether you're comfortable with the results you already have; if you're not, please tell me specifically how I may be of help to you in this process.

Answer 21

i'm comfortable with what we have until this point, but the actual solving of the systems are the reason i'm here for help

Answer 22

Which method or methods have you used in the past to solve systems of linear equations?

Answer 23

use c to solve for A or B in eq2 and use that to solve for B or D in eq3 and then solve for the other unknown. but i find this very difficult to do

Answer 24

that's the "substitution" method.
I think you have enuf info to find all four coeff A, B, C, D.

I have your four equations here:
A+B=0
3A+2B+C=0
3A +B + D = 3
C =1

Let C=1 in the 2nd equation; you'll then have

A+B = 0

Answer 25

3A+2B+1=0
3A+B+D=3

Answer 26

Now you have 3 equations in 3 unknowns and are thus able to solve for all 3 unknowns. What would you do next towards that goal?

Answer 27

solve for a in eq2 to get A=(-1-2B)/3

Answer 28

Again, that's the substitution method, and should work fine. Other methods include addition/subtraction, matrices and determinants. Can you solve this problem for A, B and D now? I'll stay with y ou as long as it takes you to solve this system.

Answer 29

i don't know any of the other methods

Answer 30

I did the problem on my calc (a TI-84) and obtained A=-1, B=1, C=1, D=5 (for reference). Why not start with A+B=0, which can be solved for either A or B:
let's let B=-A.
Substitute -A for every occurrence of B. You'll end up with 2 equations in A and D.

Answer 31

i thought that was the substitution method?

Answer 32

It's just easier to solve the first equation for B: A+B=0 results in B=-A.

Answer 33

We have used substitution, and can continue to use it.

Answer 34

Let me know how I may continue to help you. If y ou can solve this on your own, I'd prefer that you do that, for the experience. If you like, we could later discuss other methods of solving systems of linear equations.

Answer 35

I'd like to get off my computer soon.

Answer 36

i got A=-1,B=1,C= and D=2

Answer 37

I think you meant C=1.

Answer 38

yes, i missed the key

Answer 39

I agree, except that I got D=5.
You might want to check your calculation of D.

Answer 40

D=-3A-B
=-3(-1)-(1)
=2

Answer 41

going back to near the beginning, I see we had an equation that reads like this: 3 = 3A +B + D. What aer your values for A and B? Subst. them into this equation and seewhat D value you obtain.

Answer 42

D=5, i see where i went wrong thanks

Answer 43

So: are you completely comfortable with this process? Did I spend too much time on the early part of this discussion? Do you want to discuss other methods of solving systems of linear equations (not now, but later)?

Answer 44

i think i've got it, i'll try some more examples and if i have any problems i'll come back laer

Answer 45

Nice working with you! Best of luck. See you later.