I am trying to get the polynomial;

x^7+2x^6+3x^5+4x^4+6x^3+6x^2+7x+8
divided by;
x^5-2x^4-3x^3+5x^2+2x-3

into a partial fraction form so that I can create a system of equations to solve for. So far, after dividing I have
(x+1)(x-1)^2, and (x^2-x-3) as factors along with an remainder ype thing of x^2+4x+14. This gives me
A/(x+1)+ B/(x-1)+ C/(x-1)^2+ Dx+E/(x^2-x-3)
I don't know if this is right, and need help from here.
Thanks

Bobpursley already answered your question. What is it that you do not understand about his answer?

Also -- please don't keep posting the same question -- even under different names.

I don't understand how to convert the partial fractions into a system

Jo or John

very messy problem
your denominator is factored correctly
your answer to your long division is indeed x^2 + 4x + 14 but you don't say what that remainder is, because it is the remainder that you have to work with.

so you are working with

remainder/[(x+1)(x-1)^2(x^2-x-3)
= A/(x+1)+ B/(x-1)+ (Cx+D)/(x-1)^2+ (Ex+F)/(x^2-x-3)

notice I changed the numerators of your last two fractions, because they have a quadratic denominator.

I now want you to watch this video

(Broken Link Removed)

beginning at time 16:00 he starts a problem very similar to yours

To find the partial fraction decomposition of the given rational function, you're on the right track. Let's continue from where you left off.

You correctly factored the denominator:
x^5 - 2x^4 - 3x^3 + 5x^2 + 2x - 3 = (x+1)(x-1)^2(x^2-x-3)

Now, the next step is to decompose the rational function into partial fractions. The general form of the partial fraction decomposition is:

A/(x+1) + B/(x-1) + C/(x-1)^2 + Dx + E/(x^2-x-3)

To determine the constants A, B, C, D, and E, you need to equate the numerators of this decomposition with the corresponding parts of the original equation and simplify.

Step 1: Equate coefficients of x^7 terms:
On the left side, we have: 0x^7
On the right side, we have: 0/(x+1) + 0/(x-1) + 0/(x-1)^2 + 0x + 0/(x^2-x-3)

Step 2: Equate coefficients of x^6 terms:
On the left side, we have: 1x^6
On the right side, we have: 0/(x+1) + 0/(x-1) + 0/(x-1)^2 + 0x + 0/(x^2-x-3)

Step 3: Equate coefficients of x^5 terms:
On the left side, we have: 2x^5
On the right side, we have: A/(x+1) + B/(x-1) + C/(x-1)^2 + 0x + 0/(x^2-x-3)

Step 4: Equate coefficients of x^4 terms:
On the left side, we have: 4x^4
On the right side, we have: A/(x+1) + B/(x-1) + C/(x-1)^2 + 0x + 0/(x^2-x-3)

Step 5: Equate coefficients of x^3 terms:
On the left side, we have: 6x^3
On the right side, we have: A/(x+1) + B/(x-1) + C/(x-1)^2 + Dx + 0/(x^2-x-3)

Step 6: Equate coefficients of x^2 terms:
On the left side, we have: 6x^2
On the right side, we have: A/(x+1) + B/(x-1) + C/(x-1)^2 + Dx + E/(x^2-x-3)

Step 7: Equate coefficients of x terms:
On the left side, we have: 7x
On the right side, we have: A/(x+1) + B/(x-1) + C/(x-1)^2 + Dx + E/(x^2-x-3)

Step 8: Equate constant terms:
On the left side, we have: 8
On the right side, we have: A/(x+1) + B/(x-1) + C/(x-1)^2 + Dx + E/(x^2-x-3)

Now, you have obtained a system of equations:
0 = 0/(x+1) + 0/(x-1) + 0/(x-1)^2 + 0x + 0/(x^2-x-3)
1 = A/(x+1) + B/(x-1) + C/(x-1)^2 + 0x + 0/(x^2-x-3)
2 = A/(x+1) + B/(x-1) + C/(x-1)^2 + 0x + 0/(x^2-x-3)
4 = A/(x+1) + B/(x-1) + C/(x-1)^2 + 0x + 0/(x^2-x-3)
6 = A/(x+1) + B/(x-1) + C/(x-1)^2 + Dx + 0/(x^2-x-3)
6 = A/(x+1) + B/(x-1) + C/(x-1)^2 + Dx + E/(x^2-x-3)
7 = A/(x+1) + B/(x-1) + C/(x-1)^2 + Dx + E/(x^2-x-3)
8 = A/(x+1) + B/(x-1) + C/(x-1)^2 + Dx + E/(x^2-x-3)

To solve this system, you can multiply through by the common denominator (x+1)(x-1)^2(x^2-x-3) to eliminate the denominators. Then, you can combine like terms and solve for the coefficients.

At this point, you have reached the correct form for the partial fraction decomposition, and solving the system will give you the values of A, B, C, D, and E.