Methods Resolve the following: (B) x-1. (x+2)(x-3). Solution 1

We assume the expression can be written as:

$$\frac{x-1}{(x+2)(x-3)} = \frac{A}{x+2} + \frac{B}{x-3}$$

Multiply both sides by $(x+2)(x-3)$ to clear the denominators:

$$x-1 = A(x-3) + B(x+2)$$

Expand the right side:

$$x-1 = Ax - 3A + Bx + 2B$$

Group terms by powers of $x$:

$$x-1 = (A+B)x + (-3A+2B)$$

Equating the coefficients of $x$ and the constant terms on both sides, we get:

$$A+B = 1 \quad \text{(i)}$$ $$-3A+2B = -1 \quad \text{(ii)}$$

From equation (i), express $A$ in terms of $B$:

$$A = 1-B \quad \text{(iii)}$$

Substitute equation (iii) into equation (ii):

$$-3(1-B) + 2B = -1$$ $$-3 + 3B + 2B = -1$$ $$-3 + 5B = -1$$ $$5B = -1 + 3$$ $$5B = 2$$ $$B = \frac{2}{5}$$

Substitute the value of $B$ back into equation (iii) to find $A$:

$$A = 1 - \frac{2}{5}$$ $$A = \frac{5}{5} - \frac{2}{5}$$ $$A = \frac{3}{5}$$

Substitute the values of $A$ and $B$ back into the partial fraction form:

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

This can be written more cleanly as:

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