Solve for C from the Partial Fraction Solve for C from $$\dfrac{6x^2 + x - 37}{(x - 3)(x + 2)(x - 1)} = \dfrac{A}{x - 3} + \dfrac{B}{x + 2} + \dfrac{C}{x - 1}$$ A. 2 C. 4 B. 3 D. 5 Log in or register to post comments Solution Click here to… Jhun Vert Fri, 07/12/2024 - 23:50 Solution Click here to expand or collapse this section $\dfrac{6x^2 + x - 37}{(x - 3)(x + 2)(x - 1)} = \dfrac{A}{x - 3} + \dfrac{B}{x + 2} + \dfrac{C}{x - 1}$ Multiply both sides by $(x - 3)(x + 2)(x - 1)$ $6x^2 + x - 37 = A(x + 2)(x - 1) + B(x - 3)(x - 1) + C(x - 3)(x + 2)$ Set $x = 1$ $6 + 1 - 37 = C(-2)(3)$ $C = 5$ Log in or register to post comments
Solution Click here to… Jhun Vert Fri, 07/12/2024 - 23:50 Solution Click here to expand or collapse this section $\dfrac{6x^2 + x - 37}{(x - 3)(x + 2)(x - 1)} = \dfrac{A}{x - 3} + \dfrac{B}{x + 2} + \dfrac{C}{x - 1}$ Multiply both sides by $(x - 3)(x + 2)(x - 1)$ $6x^2 + x - 37 = A(x + 2)(x - 1) + B(x - 3)(x - 1) + C(x - 3)(x + 2)$ Set $x = 1$ $6 + 1 - 37 = C(-2)(3)$ $C = 5$ Log in or register to post comments
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