Simplification and expansion of algebraic fractions

Example

Simplify the following algebraic fraction and conclude if it is a polynomial or an algebraic fraction. $${\displaystyle \frac{x^2-1}{x^2-2x+1}}$$ We factorize the numerator and the denominator:

$x^2-1=(x+1)\cdot (x-1)$

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

We see that the numerator and the denominator have a common factor, therefore:

$${\displaystyle \frac{x^2-1}{x^2-2x+1}}={\displaystyle \frac{(x+1)(x-1)}{(x-1{)}^2}}={\displaystyle \frac{x+1}{x-1}}$$ And the result is an algebraic fraction.

Example

Simplify the following algebraic fraction and conclude if it is a polynomial or an algebraic fraction. $${\displaystyle \frac{x^2-4x+4}{x-2}}$$ We factorize the numerator and the denominator:

$x^2-4x+4=(x-2{)}^2$

$x-2$

We see that the numerator and the denominator have a common factor, therefore:

$${\displaystyle \frac{x^2-4x+4}{x-2}}={\displaystyle \frac{(x-2{)}^2}{x-2}}=x-2$$ And the result is a polynomial.

Example

Expand the following algebraic fraction ${\displaystyle \frac{x-1}{x+2}}$ in such a way that it has a polynomial with a root at $x=-1$ in the denominator.

It is enough to multiply the algebraic fraction by the expression $x+1$ both in the numerator and in the denominator: $${\displaystyle \frac{x-1}{x+2}}={\displaystyle \frac{x-1}{x+2}}\cdot {\displaystyle \frac{x+1}{x+1}}$$ Now, we can expand the polynomials: $${\displaystyle \frac{x-1}{x+2}}\cdot {\displaystyle \frac{x+1}{x+1}}={\displaystyle \frac{(x-1)(x+1)}{(x+2)(x+1)}}={\displaystyle \frac{x^2-1}{x\cdot (x+1)+2\cdot (x+1)}}=$$

$$={\displaystyle \frac{x^2-1}{x^2+3x+2}}$$ The result is an algebraic fraction equivalent to the initial.

Example

Expand the following algebraic fraction ${\displaystyle \frac{x+2}{x-2}}$ in such a way that it has a polynomial with a root at $x=3$ in the denominator.

It is enough to multiply the algebraic fraction by the expression $x-3$ both in the numerator and in the denominator: $${\displaystyle \frac{x+2}{x-2}}={\displaystyle \frac{x+2}{x-2}}\cdot {\displaystyle \frac{x-3}{x-3}}$$ Now, we can expand the polynomials: $${\displaystyle \frac{x+2}{x-2}}\cdot {\displaystyle \frac{x-3}{x-3}}={\displaystyle \frac{(x+2)(x-3)}{(x-2)(x-3)}}={\displaystyle \frac{x\cdot (x-3)+2\cdot (x-3)}{x\cdot (x-3)-2\cdot (x-3)}}=$$

$$={\displaystyle \frac{x^2-x-6}{x^2-5x+6}}$$ The result is an algebraic fraction equivalent to the initial.