Consider the algebraic fractions $$\dfrac{x+3}{x^2-9}$$ and $$\dfrac{x^2}{(x^2+2)\cdot(x+3)}$$, find two equivalent algebraic fractions with common denominator.
Development:
It is enough to use the steps:
- Factorization:
$$x^2-9=(x+3)\cdot(x-3)$$
$$(x^2+2)\cdot(x+3)$$
- Compute the least common multiple (l.c.m). of the polynomials in the denominators:
$$lcm\{x^2-9,(x^2+2)\cdot(x+3)\}=lcm\{(x-3)\cdot(x+3),(x^2+2)\cdot(x+3)\}=$$
$$=(x^2+2)\cdot(x+3)\cdot(x-3)$$
- We divide the l.c.m. by every denominator and multiply it by the respective numerator. The result is the numerator of the algebraic fraction; the denominator is the l.c.m.
$$\dfrac{(x^2+2)\cdot(x+3)\cdot(x-3)}{x^2-9}=x^2+2 \Rightarrow (x+3)\cdot(x^2-9)=x\cdot(x^2-9)+3\cdot(x^2-9)=$$
$$=x^3+3x^2-9x-27 \Rightarrow \dfrac{x^3+3x^2-9x-27}{(x^2+2)\cdot(x+3)\cdot(x-3)}$$
$$\dfrac{(x^2+2)\cdot(x+3)\cdot(x-3)}{(x^2+2)(x+3)}=x-3 \Rightarrow x^2\cdot(x-3)=x^3-3x^2 \Rightarrow $$
$$\Rightarrow \dfrac{x^3-3x^2}{(x^2+2)\cdot(x+3)\cdot(x-3)}$$
Solution:
$$\dfrac{x^3+3x^2-9x-27}{(x^2+2)\cdot(x+3)\cdot(x-3)}$$ and $$\dfrac{x^3-3x^2}{(x^2+2)\cdot(x+3)\cdot(x-3)}$$