(a) Express \(\frac{2x^{2} - 5x + 1}{x^{3} - 4x^{2} + 3x}\) in partial fractions. (b)

Explanation

(a) \(\frac{2x^{2} - 5x + 1}{x^{3} - 4x^{2} + 3x}\)

\(x^{3} - 4x^{2} + 3x = x(x^{2} - 4x + 3)\)

\(x^{2} - 4x + 3 = (x - 1)(x - 3)\)

\(\therefore x^{3} - 4x^{2} + 3x = x(x - 1)(x - 3)\)

\(\implies \frac{2x^{2} - 5x + 1}{x^{3} - 4x^{2} + 3x} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x - 3}\)

\(\implies \frac{2x^{2} - 5x + 1}{x^{3} - 4x^{2} + 3x} = \frac{A(x - 1)(x - 3) + B(x)(x - 3) + C(x)(x - 1)}{x(x - 1)(x - 3)}\)

Equating, we have

\(2x^{2} - 5x + 1 = A(x - 1)(x - 3) + B(x)(x - 3) + C(x)(x - 1) \)

Let x = 1, we have

\(2 - 5 + 1 = -2B \implies B = 1\)

Let x = 3, we have

\(18 - 15 + 1 = 6C \implies C = \frac{4}{6} = \frac{2}{3}\)

Let x = 0, we have

\(0 - 0 + 1 = 3A \implies A = \frac{1}{3}\)

\(\therefore \frac{2x^{2} - 5x + 1}{x^{3} - 4x^{2} + 3x} = \frac{1}{3x} + \frac{1}{x - 1} + \frac{2}{3(x - 3)}\)

(b) \(\begin{vmatrix} x - 3 & -4 & 3 \\ 5 & 2 & 2 \\ 2 & -4 & 6 - x \end{vmatrix} = -24\)

\(\implies (x - 3)(12 - 2x + 8) - (-4)(30 - 5x - 4) + 3(-20 - 4) = -24\)

\((x - 3)(20 - 2x) + 4(26 - 5x) - 72 + 24 = 0\)

\(20x - 2x^{2} - 60 + 6x + 104 - 20x - 48 = 0\)

\(-2x^{2} + 6x - 4 = 0\)

\(2x^{2} - 6x + 4 = 0\)

\(2x^{2} - 4x - 2x + 4 = 0 \implies 2x(x - 2) - 2(x - 2) = 0\)

\((2x - 2)(x - 2) = 0 \implies \text{x = 1 or 2}\).