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

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

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Answer 1 · 2016-10-01T01:00:26

$3 \frac{x^{2} + x + 4}{(x - 5) (x^{2} + 9)} = \frac{3}{x^{2} + 9} + \frac{3}{x - 5}$ $3(x^2 + x + 4)/((x - 5)(x^2 + 9)) = 3 /(x^2+ 9) + 3/(x - 5)$

Explanation:

$3 \frac{x^{2} + x + 4}{(x - 5) (x^{2} + 9)} = \frac{A + B x}{x^{2} + 9} + \frac{C}{x - 5}$ $3(x^2 + x + 4)/((x - 5)(x^2 + 9)) = (A + Bx)/(x^2+ 9) + C/(x - 5)$

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

Let x = 5:

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

$C = 3$ $C = 3$

Let x = 0:

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

$12 = - 5 A + 27$ $12 = -5A + 27$

$A = 3$ $A = 3$

Let x = 1:

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

$18 = - 12 - 4 B + 30$ $18 = -12 - 4B + 30$

$B = 0$ $B = 0$

$3 \frac{x^{2} + x + 4}{(x - 5) (x^{2} + 9)} = \frac{3}{x^{2} + 9} + \frac{3}{x - 5}$ $3(x^2 + x + 4)/((x - 5)(x^2 + 9)) = 3 /(x^2+ 9) + 3/(x - 5)$

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

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Explanation:

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How do you express [(3x2)+3x+12(x−5)(x2+9)] [((3x^2)+3x+12) / ((x-5)(x^2+9))] in partial fractions?

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Explanation:

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