Question

What are the respective values of A,B and C, so that x-6/x³+2x =A/x +Bx+C/x²+2

Answer 1

To find the values of \( A \), \( B \), and \( C \) such that

\[
\frac{x - 6}{x^3 + 2x} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^2 + 2},
\]

we first simplify the left-hand side.

We can factor the denominator of the left side:

\[
x^3 + 2x = x(x^2 + 2).
\]

So we have:

\[
\frac{x - 6}{x(x^2 + 2)}.
\]

Next, rewrite the right-hand side with a common denominator:

The common denominator for the right-hand side is \( x^2(x^2 + 2) \). Thus, we rewrite the fractions:

\[
\frac{A}{x} = \frac{A \cdot x(x^2 + 2)}{x^2(x^2 + 2)} = \frac{Ax(x^2 + 2)}{x^2(x^2 + 2)},
\]
\[
\frac{B}{x^2} = \frac{B \cdot (x^2 + 2)}{x^2(x^2 + 2)} = \frac{B(x^2 + 2)}{x^2(x^2 + 2)},
\]
\[
\frac{C}{x^2 + 2} = \frac{C \cdot x^2}{x^2(x^2 + 2)} = \frac{Cx^2}{x^2(x^2 + 2)}.
\]

Now combine these fractions:

\[
\frac{Ax(x^2 + 2) + B(x^2 + 2) + Cx^2}{x^2(x^2 + 2)}.
\]

Equating both sides, we have:

\[
\frac{x - 6}{x(x^2 + 2)} = \frac{Ax(x^2 + 2) + B(x^2 + 2) + Cx^2}{x^2(x^2 + 2)}.
\]

Multiplying both sides by \( x^2(x^2 + 2) \) leads to:

\[
x - 6 = Ax(x^2 + 2) + B(x^2 + 2) + Cx^2.
\]

We now expand the right side:

\[
Ax^3 + 2Ax + Bx^2 + 2B + Cx^2 = Ax^3 + (B + C)x^2 + 2Ax + 2B.
\]

We can now set coefficients from both sides of the equation. For the left side \( x - 6 \), we have:

1. Coefficient of \( x^3 \): \( A = 0 \)
2. Coefficient of \( x^2 \): \( B + C = 0 \)
3. Coefficient of \( x \): \( 2A = 1 \) (since \( A = 0 \) this doesn't apply)
4. Constant term: \( 2B = -6 \)

From \( 2B = -6 \):

\[
B = -3.
\]

Using \( B + C = 0 \):

\[
-3 + C = 0 \Rightarrow C = 3.
\]

Now, substituting back into our equation, we have:

\[
A = 0, \quad B = -3, \quad C = 3.
\]

Thus, the values of \( A \), \( B \), and \( C \) are:

\[
\boxed{(0, -3, 3)}.
\]