(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.
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}).