How do you express $[((3x^2)+3x+12) / ((x-5)(x^2+9))]$ in partial fractions?

Question

1450 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-01T01:00:26

$3(x^2 + x + 4)/((x - 5)(x^2 + 9)) = 3 /(x^2+ 9) + 3/(x - 5)$

$3(x^2 + x + 4)/((x - 5)(x^2 + 9)) = (A + Bx)/(x^2+ 9) + C/(x - 5)$

$3(x^2 + x + 4) = (A + Bx)(x - 5) + C(x^2+ 9)$

Let x = 5:

$3(5^2 + 5 + 4) = C(5^2+ 9)$

$C = 3$

Let x = 0:

$3(0^2 + 0 + 4) = (A + B0)(0 - 5) + 3(0^2+ 9)$

$12 = -5A + 27$

$A = 3$

Let x = 1:

$3(1^2 + 1 + 4) = (3 + B)(1 - 5) + 3(1^2+ 9)$

$18 = -12 - 4B + 30$

$B = 0$

$3(x^2 + x + 4)/((x - 5)(x^2 + 9)) = 3 /(x^2+ 9) + 3/(x - 5)$

How do you express [((3x^2)+3x+12) / ((x-5)(x^2+9))] [((3x^2)+3x+12) / ((x-5)(x^2+9))] in partial fractions?