# What are the partial fractions of (2x + 1)/ (-x^3 - 6x^2 + 13x + 42)?

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### 1 Answer

We have to find the partial fractions of (2x + 1)/ (-x^3 - 6x^2 + 13x + 42)

We first need to factorize the denominator (-x^3 - 6x^2 + 13x + 42)

-x^3 - 2x^2 - 4x^2 - 8x + 21x + 42

=> -x^2(x + 2) - 4x( x + 2) + 21(x + 2)

=> (x + 2)((-x^2 - 4x + 21)

=> (x + 2)((-x^2 - 7x + 3x + 21)

=> (x + 2)((-x(x + 7) + 3(x + 7))

=> (x + 2)(3 - x)(7 + x)

The partial fraction is going to be of the form (2x + 1)/(x + 2)(3 - x)(7 + x) = A/(x + 2) + B/(3 - x) + C/(7 + x)

Now to find the value of A, B and C follow this easy method. Let's determine A first.

Write (2x + 1)/(x + 2)(3 - x)(7 + x) , as we are finding the coefficient for x + 2, cover that and substitute all the values of x with the root given by x + 2 = 0 or x = -2.

=> [2*(-2) + 1]/(3 + 2)(7 - 2) = -3/25

Similarly for B, replace x = 3 in (2x + 1)/(x + 2)(7 + x)

=> B = 7/50

For C, replace x = -7 in (2x + 1)/(x + 2)(3 - x)

=> C = 13/ 50

**The partial fractions of (2x + 1)/(x + 2)(3 - x)(7 + x) are: **

**(-3/25)/(x + 2) + (7/50)/(3 - x) +(13/50)/(7 + x)**