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I am trying to solve for $a$, $b$, and $c$ in the expression below, but I have found that the way I tried to solve it is convoluted and did not work out. I believed that by solving for x, I would be able to solve for the coefficients.

(1) The original expression: $$\frac{x^{2} - x + 3}{(x^2 + 2)(2x - 1)} = \frac{ax + b}{x^2 + 2} + \frac{c}{2x-1}$$

(2) Multiply both sides by $(x^2 + 2)(2x - 1)$.

$$x^{2} - x + 3 = (ax + b)(2x - 1)+ c(x^2 + 2)$$

(3) Simplify.

$$x^{2} - x + 3 = 2ax^{2} -ax + 2bx - b + cx^{2} + 2c$$

(4) Subtract both sides by $2ax^{2} -ax + 2bx - b + cx^{2} + 2c$.

$$x^{2} - x + 3- 2ax^{2} + ax - 2bx + b - cx^{2} - 2c = 0$$

(5) Arrange into quadratic form.

$$(1 - 2a - c)x^{2} + (a - 2b - 1)x + (- 2c + b + 3) = 0$$

(6) Apply the quadratic formula.

$$x = \frac{-(a - 2b - 1) \pm \sqrt{(a - 2b - 1)^{2} - 4(1 - 2a - c)(- 2c + b + 3))}}{2(1 - 2a - c)}$$

(7) Attempt to simplify.

$$\frac{-a + 2b + 1 \pm \sqrt{a^{2} + 4ab + 12a + 4b^{2} + 20c - 12 - 4c^{2} + 4bc}}{2 - 4a - 2c}$$

(8) Ask for help.

Galen
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2 Answers2

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From step (3) you can conclude that: $$ 2a + c = 1 $$

$$ -a+2b = -1 $$

$$ -b+2c = 3 $$

So there are three equations with 3 unknowns.

user153245
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You have to solve this by using a system of equations. For starters, take step (2) and substitute $x=0.$ Then use another value of $x$ to get another equation and so on. For more see https://youtu.be/hnIs8uwAT3U.

John Molokach
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