Consider the fractions $$\dfrac{x+3}{x-1}$$ and $$\dfrac{x-3}{x+1}$$, find an expansion in such a way that the denominators have a root at $$x=-2$$ and $$x=4$$.
Development:
For the denominator to have a root at $$x=-2$$, it is enough to multiply the algebraic fraction by the expression $$x+2$$, both in the numerator and in the denominator:
$$$\dfrac{x+3}{x-1}\cdot\dfrac{x+2}{x+2}=\dfrac{(x+3)\cdot(x+2)}{(x-1)\cdot(x+2)}=\dfrac{x\cdot(x+2)+3\cdot(x+2)}{x\cdot(x+2)-1\cdot(x+2)}=\dfrac{x^2+5x+6}{x^2+x+4}$$$
For the denominator to have a root at $$x=4$$, it is enough to multiply the algebraic fraction by the expression $$x-4$$, both in the numerator and in the denominator:
$$$\dfrac{x-3}{x+1}\cdot\dfrac{x-4}{x-4}=\dfrac{(x-3)\cdot(x-4)}{(x+1)\cdot(x-4)}=\dfrac{x\cdot(x-4)-3\cdot(x-4)}{x\cdot(x-4)+1\cdot(x-4)}=\dfrac{x^2-7x+12}{x^2-3x-4}$$$
Solution:
$$\dfrac{x^2+5x+6}{x^2+x+4}$$
$$\dfrac{x^2-7x+12}{x^2-3x-4}$$