(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 \en...

Answer Details

(a) Partial fractions of \(\dfrac{2x^2-5x+1}{x^3-4x^2+3x}.\)

Factorise the denominator: \(x^3-4x^2+3x=x(x^2-4x+3)=x(x-1)(x-3).\)

Write

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

Multiply out: \(2x^2-5x+1=A(x-1)(x-3)+Bx(x-3)+Cx(x-1).\)

\(x=0:\ 1=A(-1)(-3)=3A\Rightarrow A=\dfrac{1}{3}.\)

\(x=1:\ 2-5+1=-2=B(1)(-2)=-2B\Rightarrow B=1.\)

\(x=3:\ 18-15+1=4=C(3)(2)=6C\Rightarrow C=\dfrac{2}{3}.\)

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

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

Expand along the first row:

\[(x-3)\big[2(6-x)-2(-4)\big]-(-4)\big[5(6-x)-2(2)\big]+3\big[5(-4)-2(2)\big].\]

\[=(x-3)(20-2x)+4(26-5x)+3(-24).\]

Now \((x-3)(20-2x)=-2x^2+26x-60,\) so the total is

\[-2x^2+26x-60+104-20x-72=-2x^2+6x-28.\]

Set equal to \(-24:\)

\[-2x^2+6x-28=-24\Rightarrow -2x^2+6x-4=0\Rightarrow x^2-3x+2=0.\]

\[(x-1)(x-2)=0\Rightarrow x=1\ \text{or}\ x=2.\]