(a) Express \(\frac{2x^{2} - 5x + 1}{x^{3} - 4x^{2} + 3x}\) in partial fractions.
(b) If \(\begin{vmatrix} x - 3 & -4 & 3 \\ 5 & 2 & 2 \\ 2 & -4 & 6 - x \end{vmatrix} = -24\), find the value of x.
(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) \(\begin{vmatrix} x - 3 & -4 & 3 \\ 5 & 2 & 2 \\ 2 & -4 & 6 - x \end{vmatrix} = -24\)
\(\implies (x - 3)(12 - 2x + 8) - (-4)(30 - 5x - 4) + 3(-20 - 4) = -24\)
\((x - 3)(20 - 2x) + 4(26 - 5x) - 72 + 24 = 0\)
\(20x - 2x^{2} - 60 + 6x + 104 - 20x - 48 = 0\)
\(-2x^{2} + 6x - 4 = 0\)
\(2x^{2} - 6x + 4 = 0\)
\(2x^{2} - 4x - 2x + 4 = 0 \implies 2x(x - 2) - 2(x - 2) = 0\)
\((2x - 2)(x - 2) = 0 \implies \text{x = 1 or 2}\).
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