Resolve frac{x^2 + 13x + 15}{(2x + 3)(x + 3)^ | vedclass

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(A) Let $\frac{x^2 + 13x + 15}{(2x + 3)(x + 3)^2} = \frac{A}{2x + 3} + \frac{B}{x + 3} + \frac{C}{(x + 3)^2}$.Multiplying both sides by $(2x + 3)(x +

Vedclass · Accepted Answer

$\frac{1}{x + 3} - \frac{1}{2x + 3} + \frac{5}{(x + 3)^2}$

Vedclass · Answer

(A) Let $\frac{x^2 + 13x + 15}{(2x + 3)(x + 3)^2} = \frac{A}{2x + 3} + \frac{B}{x + 3} + \frac{C}{(x + 3)^2}$.Multiplying both sides by $(2x + 3)(x + 3)^2$,we get:$x^2 + 13x + 15 = A(x + 3)^2 + B(2x + 3)(x + 3) + C(2x + 3)$.Setting $x = -3$: $(-3)^2 + 13(-3) + 15 = C(2(-3) + 3) \implies 9 - 39 + 15 = -3C \implies -15 = -3C \implies C = 5$.Setting $x = -\frac{3}{2}$: $(-\frac{3}{2})^2 + 13(-\frac{3}{2}) + 15 = A(-\frac{3}{2} + 3)^2 \implies \frac{9}{4} - \frac{39}{2} + 15 = A(\frac{3}{2})^2 \implies \frac{9 - 78 + 60}{4} = A(\frac{9}{4}) \implies -9 = 9A \implies A = -1$.Comparing coefficients of $x^2$: $1 = A + 2B \implies 1 = -1 + 2B \implies 2 = 2B \implies B = 1$.Thus,the partial fractions are $\frac{1}{x + 3} - \frac{1}{2x + 3} + \frac{5}{(x + 3)^2}$.

Resolve $\frac{x^2 + 13x + 15}{(2x + 3)(x + 3)^2}$ into partial fractions.

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