If frac{ax^2 + bx + c}{(x - 1)(x + 2)(2x + 3) | vedclass

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(A) Given $\frac{ax^2 + bx + c}{(x - 1)(x + 2)(2x + 3)} = \frac{3}{x - 1} + \frac{2}{x + 2} - \frac{5}{2x + 3}$.Multiplying both sides by the denomina

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$a = 5$

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(A) Given $\frac{ax^2 + bx + c}{(x - 1)(x + 2)(2x + 3)} = \frac{3}{x - 1} + \frac{2}{x + 2} - \frac{5}{2x + 3}$.Multiplying both sides by the denominator $(x - 1)(x + 2)(2x + 3)$,we get:$ax^2 + bx + c = 3(x + 2)(2x + 3) + 2(x - 1)(2x + 3) - 5(x - 1)(x + 2)$.Expanding the terms:$3(2x^2 + 7x + 6) + 2(2x^2 + x - 3) - 5(x^2 + x - 2)$$= (6x^2 + 21x + 18) + (4x^2 + 2x - 6) - (5x^2 + 5x - 10)$$= (6 + 4 - 5)x^2 + (21 + 2 - 5)x + (18 - 6 + 10)$$= 5x^2 + 18x + 22$.Comparing coefficients,we get $a = 5$,$b = 18$,and $c = 22$.Thus,$a = 5$ is correct.

If $\frac{ax^2 + bx + c}{(x - 1)(x + 2)(2x + 3)} = \frac{3}{x - 1} + \frac{2}{x + 2} - \frac{5}{2x + 3}$,then:

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