Solve the partial fraction (X-1)(X+1)/(X+2)(X^2-3X+1).

Answer 1

To solve the partial fraction decomposition, we first factor the denominator:

X^2 - 3X + 1 = (X - φ)(X - ψ)

where φ and ψ represent the roots of the quadratic equation.

The roots of the quadratic equation can be found using the quadratic formula:

X = (-(-3) ± √((-3)^2 - 4(1)(1))) / (2(1))
= (3 ± √(5)) / 2

So, the roots of the quadratic equation are (3 + √5) / 2 and (3 - √5) / 2.

Therefore, the factored form of the denominator is:

(X + 2) (X - (3 + √5) / 2) (X - (3 - √5) / 2)

Now, let's write the partial fraction decomposition:

(X - 1)(X + 1) / ((X + 2)(X^2 - 3X + 1))
= A / (X + 2) + (B(X - (3 + √5) / 2) + C(X - (3 - √5) / 2)) / (X^2 - 3X + 1)

To find the values of A, B, and C, we can multiply both sides of the equation by the denominator:

(X - 1)(X + 1) = A(X^2 - 3X + 1) + (B(X - (3 + √5) / 2) + C(X - (3 - √5) / 2))(X + 2)

Expanding the equation and combining like terms, we get:

X^2 - 1 = A X^2 + (-3A + B + C) X + (A + (2B - (3 + √5) / 2) + (2C - (3 - √5) / 2))

Now, we need to match the coefficients of the corresponding powers of X on both sides of the equation.

For X^2:
A = 1

For X^1:
-3A + B + C = 0

For X^0:
A + (2B - (3 + √5) / 2) + (2C - (3 - √5) / 2) = -1

Simplifying the equations, we have:

A = 1

-3 + B + C = 0

2B + 2C - 3 + √5 - 3 - √5 = -1

Simplifying further:

B + C = 3

2B + 2C - 6 = -1

2B + 2C = 5

Simplifying again:

B + C = 3

B + C = 2.5

We have two conflicting equations, which means there is no solution for this partial fraction decomposition.

Answer 2

(x-1)(x+1)/((x+2)(x^2-3x+1))
= 3/(11(x+2)) + (8x-7)/(11(x^2-3x+1))

Answer 3

I apologize for the mistake in my previous response. You are correct, the partial fraction decomposition of (x-1)(x+1)/((x+2)(x^2-3x+1)) can be written as:

(x-1)(x+1)/((x+2)(x^2-3x+1)) = 3/(11(x+2)) + (8x-7)/(11(x^2-3x+1))

This can be obtained by finding the values of A, B, and C through a similar process as previously described.