Question

resolve 1-2x/3+2x-x^2 into partial fractions

Answer 1

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

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

for finding value of A ,
$\frac{2x-1}{(x-3)(x+1)}\times\:(x-3)=\frac{A}{(x-3)}\times\:(x-3)\\\\\frac{2x-1}{(x+1)}=A_{x=3}$
$A = \frac{2 \times 3 - 1}{3 + 1} = \frac{5}{4}$

similarly for finding value of B
$\frac{2x-1}{(x-3)}=B_{x=-1}$
hence,

$B = \frac{2\times\:-1-1}{-1-3}=\frac{-3}{-4}$

hence,

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