# Write as partial fractions : [5x^2 + 10 x + 3] / [ x^3 + 2x^2 – x – 2]

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We have to express [5x^2 + 10 x + 3] / [x^3 + 2x^2 – x – 2] as partial fractions.

First, we factorize the denominator

[5x^2 + 10 x + 3] / [x^3 + 2x^2 – x – 2]

=> [5x^2 + 10 x + 3] / [x^2(x + 2)-1(x + 2]

=> [5x^2 + 10 x + 3] / [(x^2 – 1) (x + 2)]

=> [5x^2 + 10 x + 3] / [(x – 1) (x + 1) (x + 2)]

Let us write the given expression as A / (x – 1) +B /(x + 1) + C/(x + 2)

=> [A(x +1) (x +2) + B(x – 1) (x +2) + C(x – 1) (x +1)]/ [(x – 1) (x + 1) (x + 2)]

=> [A(x^2 + 3x + 2) + B(x^2 + x – 2) + C(x^2 – 1)] / [(x – 1) (x + 1) (x + 2)]

[A(x^2 + 3x + 2) + B(x^2 + x – 2) + C(x^2 – 1)] / [(x – 1) (x + 1) (x + 2)] = [5x^2 + 10 x + 3] / [(x – 1) (x + 1) (x + 2)]

=> [A(x^2 + 3x + 2) + B(x^2 + x – 2) + C(x^2 – 1)] = [5x^2 + 10 x + 3]

=> Ax^2 + 3Ax + 2A + Bx^2 + Bx – 2B + Cx^2 – C = 5x^2 + 10 x + 3

Equating the coefficients of x^2, x and the numeric term we get:

(A + B + C) = 5 … (1)

(3A + B) = 10 … (2)

(2A – 2B – C) = 3… (3)

Add (1) and (3)

=> 3A – B = 8

Now 3A – B = 8 and 3A + B = 10

=> 6A = 18

=> A = 3

B = 3A – 8 = 9 – 8 = 1

C = 5 – A – B = 5 – 3 – 1 = 1

Therefore we get A / (x – 1) +B /(x + 1) + C/(x + 2) = 3 / (x – 1) +1 /(x + 1) + 1/(x + 2)

The given expression written as partial fractions is:

**3/(x – 1) +1/(x + 1) + 1/(x + 2)**