A partial fractions question

cjsage52

I need help with this question:

Express this as a partial fraction

5x^3+7
(x-3)(x^2+1)

anyone?

(x - 3)(x^2 + 1) = x^3 - 3x^2 + x - 3
(5x^3 + 7)/[(x - 3)(x^2 + 1)] = 5 + (15x^2 - 5x + 22)/[(x - 3)(x^2 + 1)]

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The identity

(15x^2 - 5x + 22)/[(x - 3)(x^2 + 1)] = A/(x - 3) + (Bx + C)/(x^2 + 1)

is true if (15x^2 - 5x + 22) = A(x^2 + 1) + (Bx + C)(x - 3).

Take x = 3: 142 = 10A, A = 71/5.
Compare coefficients of x^2: 15 = A + B, B = 4/5.
Take x = 0: 22 = A - 3C: C = -13/5.

So

(15x^2 - 5x + 22)/[(x - 3)(x^2 + 1)] = (1/5)[71/(x - 3) + (4x - 13)/(x^2 + 1)]

(5x^3 + 7)/[(x - 3)(x^2 + 1)] = 5 + (1/5)[71/(x - 3) + (4x - 13)/(x^2 + 1)]

Reply 3

cjsage52

Awesome, Thank you so much. Now I know what is was doing wrong. I did not factorize the orignal equation. Thanks again

Reply 4

username9816

(5x^3 + 7)/[(x - 3)(x^2 + 1) = A + B/(x - 3) + (Cx + D)/(x^2 + 1) = [A(x - 3)(x^2 + 1) + B(x^2 + 1) + (Cx + D)(x - 3)]/[(x - 3)(x^2 + 1)]

Equating numerators:

-> A(x - 3)(x^2 + 1) + B(x^2 + 1) + (Cx + D)(x - 3) = 5x^3 + 7
-> A(x^3 - 3x^2 + x - 3) + Bx^2 + B + Cx^2 - 3Cx + Dx - 3D = 5x^3 + 7
-> Ax^3 - 3Ax^2 + Ax - 3A + Bx^2 + B + Cx^2 - 3Cx + Dx - 3D = 5x^3 + 7
-> Ax^3 + (B + C - 3A)x^2 + (A - 3C + D)x + (B - 3A - 3D) = 5x^3 + 7

As the LHS and RHS are identical, equate co-efficients:

-> A = 5.
-> B + C - 3A = 0 -> B + C = 15 (1)
-> A + D - 3C = 0 -> D - 3C = -5 -> 3C - D = 5 (2)
-> B - 15 - 3D = 7 -> B - 3D = 22 (3)

(1): B = 15 - C
-> Sub. Into (2): 15 - C - 3D = 22 -> -C - 3D = 7 -> C + 3D = -7 -> C = -7 - 3D
(2): 3(-7 - 3D) - D = 5 -> -3(7 + 3D) - D = 5 -> -10D - 21 = 5 -> -10D = 26 -> 10D = -26 -> D = -13/5.
Hence: C = -7 + 39/5 = 4/5
Hence: B = 15 - C = 15 - 4/5 = 71/5

Therefore: (5x^3 + 7)/[(x - 3)(x^2 + 1) = 5 + 71/[5(x - 3)] + (4x - 13)/[5(x^2 + 1)]