Express 5x^2+7x+8/(x+1)(x^2+2x+3) as a sum of partial fractions

Answer 1

To express the given expression as a sum of partial fractions, we need to break it down into simpler fractions involving the factors of the denominator. Since the denominator has factors (x+1) and (x^2+2x+3), we will have the following form for partial fractions:

5x^2 + 7x + 8 / (x+1)(x^2+2x+3) = A / (x+1) + (Bx + C) / (x^2+2x+3),

for some constants A, B, and C, which we will determine.

Multiplying both sides by the denominator (x+1)(x^2+2x+3), we get:

5x^2 + 7x + 8 = A(x^2 + 2x + 3) + (Bx + C)(x + 1).

Expanding both sides, we have:

5x^2 + 7x + 8 = Ax^2 + 2Ax + 3A + Bx^2 + Bx + Cx + C.

Now, we will equate the coefficients of like terms on both sides:

For the x^2 terms:

5 = A + B => B = 5 - A.

For the x terms:

7 = 2A + B + C => C = 7 - 2A - B.

For the constant terms:

8 = 3A + C.

Now, we have a system of equations involving A, B, and C:

(1) B = 5 - A
(2) C = 7 - 2A - B
(3) 8 = 3A + C

We will solve this system to find A, B, and C.

Substituting (1) into (2), we get:

C = 7 - 2A - (5 - A) => C = 2 - A.

Next, substitute C in terms of A from (2) into (3):

8 = 3A + (2 - A) => 8 = 2A + 2.

Solving for A, we obtain:

A = 3.

Now that we have A, we can determine B and C using (1) and (2):

B = 5 - A = 5 - 3 = 2
C = 2 - A = 2 - 3 = -1

So the original expression becomes:

5x^2 + 7x + 8 / (x+1)(x^2+2x+3) = 3 / (x+1) + (2x - 1) / (x^2+2x+3).