5x + 6/x^2 - 4 write this as a sum of partial fractions

To write the expression 5x + 6/x^2 - 4 as a sum of partial fractions, we first need to factor the denominator x^2:

5x + 6/x^2 - 4 = 5x + 6/x^2 - 4

Now, we can write the expression as a sum of partial fractions:

5x + 6/x^2 - 4 = A/x + B/x^2 + C

To find the values of A, B, and C, we need to multiply both sides by x^2:

5x^3 + 6 - 4x^2 = A(x^2) + Bx + C(x^2)

Now, we can combine like terms and equate the coefficients of the corresponding terms:

5x^3 + 6 - 4x^2 = Ax^2 + Bx + Cx^2

5x^3 + 6 - 4x^2 = (A + C)x^2 + Bx

Comparing coefficients:

A + C = -4
B = 5
0 = 6

From the third equation, we get C = -4 - A. Substituting this into the first equation, we get:

A + (-4 - A) = -4
-4 = -4

This equation is true for all values of A, so A can be any real number. Let's choose A = 0 for simplicity. Then, C = -4 - 0 = -4. Now we can substitute A = 0 and C = -4 back into the second equation to solve for B:

B = 5

Therefore, the expression 5x + 6/x^2 - 4 can be written as a sum of partial fractions as:

5x + 6/x^2 - 4 = 5 - 4/x^2

$$\frac{5x +6}{x^2 - 4} = \frac{5x+6}{(x+2)(x-2)}$$

$$=\frac{A}{x+2} + \frac{B}{x-2}$$

$$=\frac{A(x-2)+B(x+2)}{(x+2)(x-2)}$$

$$= \frac{(A+B)x+(2B-2A)}{(x+2)(x-2)}$$

Equating coefficients:

$$A+B = 5$$

$$2B-2A = 6$$

Solving these by elimination or substitution:

$$A = 2$$

$$B = 3$$

So,

$$\frac{5x +6}{x^2 - 4} = \frac{2}{x+2} + \frac{3}{x-2}$$