# Resolve the rational function (-3x^6 +9x^5 -12x^4 +18x^3 -22x^2 +14x -6)/(x^4 -3x^3 +4x^2 -3x +1) into partial fractions.

Since the order of polynomial found at numerator is larger than the order of poynomial at denominator, you need to perform the division of polynomials, using long division, such that:

`(-3x^6 +9x^5 -12x^4 +18x^3 -22x^2 +14x -6)/(x^4 -3x^3 +4x^2 -3x +1) = -3x^2 + (9x^3-19x^2+14x-6)/(x^4 -3x^3 +4x^2 -3x +1)`

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Since the order of polynomial found at numerator is larger than the order of poynomial at denominator, you need to perform the division of polynomials, using long division, such that:

`(-3x^6 +9x^5 -12x^4 +18x^3 -22x^2 +14x -6)/(x^4 -3x^3 +4x^2 -3x +1) = -3x^2 + (9x^3-19x^2+14x-6)/(x^4 -3x^3 +4x^2 -3x +1)`

You need to use the partial fraction decomposition to convert the given fraction into partial irreducible fractions, hence, you need to convert the original form of denominator into its factored form.

You should notice that `x = 1` represents a solution to the equation `x^4 -3x^3 +4x^2 -3x +1 = 0` , hence, you may write the denominator, such that:

`x^4 -3x^3 +4x^2 -3x +1 = (x - 1)(ax^3 + bx^2 + cx + d)`

`x^4 -3x^3 +4x^2 -3x +1 = ax^4 + bx^3 + cx^2 + dx - ax^3 - bx^2 - cx - d`

`x^4 -3x^3 +4x^2 -3x +1 = ax^4 + x^3(b - a) + x^2(c - b) + x(d - c) - d`

Equating the coeffcients of like powers yields:

`a = 1`

`b - a = -3 => b - 1 = -3 => b = -2`

`c - b = 4 => c = 4 + b => c = 4 - 2 => c = 2`

`-d = 1 => d = -1`

`x^4 -3x^3 +4x^2 -3x +1 = (x - 1)(x^3 - 2x^2 + 2x - 1)`

You need to convert the factor `x^3 - 2x^2 + 2x - 1` into its factored form, such that:

`x^3 - 2x^2 + 2x - 1 = (x^3 - 1) - 2x(x - 1)`

`x^3 - 2x^2 + 2x - 1 = (x - 1)(x^2 + x + 1) - 2x(x - 1)`

Factoring out `x - 1` , yields:

`x^3 - 2x^2 + 2x - 1 = (x - 1)(x^2 + x + 1 - 2x)`

`x^3 - 2x^2 + 2x - 1 = (x - 1)(x^2 - x + 1)`

Hence, converting the denominator into its factored form yields:

`x^4 -3x^3 +4x^2 -3x +1 = (x - 1)(x - 1)(x^2 - x + 1)`

The factor `x^2 - x + 1` cannot be decomposed any further.

You need to use partial fraction decomposition, such that:

`(9x^3-19x^2+14x-6)/((x - 1)^2(x^2 - x + 1)) = a/(x - 1) + b/(x - 1)^2 + (cx + d)/(x^2 - x + 1)`

`9x^3-19x^2+14x-6= a(x - 1)(x^2 - x + 1) + b(x^2 - x + 1) + (cx + d)(x^2 - 2x + 1)`

`9x^3-19x^2+14x-6= a(x^3 - x^2 + x - x^2 + x - 1) + bx^2 - bx + b + cx^3 - 2cx^2 + cx + dx^2 - 2dx + d`

`9x^3-19x^2+14x-6= ax^3 - ax^2 + ax - ax^2 + ax - a +bx^2 - bx + b + cx^3 - 2cx^2 + cx + dx^2 - 2dx + d`

Grouping the terms that contain the same power of `x` , yields:

`9x^3-19x^2+14x-6= x^3(a + c) + x^2(-a + b - 2c + d) + x(2a - b + c - 2d) - a + b + d`

Equating the coefficients of like powers yields:

`a + c = 9 => a = 9 - c`

`-a + b - 2c + d = -19 => c - 9 + b - 2c + d = -19 => -c + b + d = -10`

`2a - b + c - 2d = 14 => 18 - 2c - b + c - 2d = 14 => -c - b - 2d = -4`

`- a + b + d = -6 => c - 9 + b + d = -6`

You need to solve for `b,c,d` the following system such that:

`{(-c + b + d = -10),(-c - b - 2d = -4),(c + b + d = 3):} => -d = -1 => d = 1`

`2b + 2 = -7 => 2b = -9 => b = -9/2`

`c = 3 + 9/2 - 1 => c = 13/2`

`a = 9 - c =>a = 9 - 13/2 => a = 5/2`

`(9x^3-19x^2+14x-6)/((x - 1)^2(x^2 - x + 1)) = 5/(2(x - 1)) - 9/(2(x - 1)^2) + ((13/2)x + 1)/(x^2 - x + 1)`

Hence, decomposing the given fraction into partial fractions, yields ` (-3x^6 +9x^5 -12x^4 +18x^3 -22x^2 +14x -6)/(x^4 -3x^3 +4x^2 -3x +1) = -3x^2 + 5/(2(x - 1)) - 9/(2(x - 1)^2) + ((13/2)x + 1)/(x^2 - x + 1).`

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