Problems from Simplification and expansion of algebraic fractions

Consider the fractions $\frac{x + 3}{x - 1}$ and $\frac{x - 3}{x + 1}$ , find an expansion in such a way that the denominators have a root at $x = - 2$ and $x = 4$ .

See development and solution

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:

$\frac{x + 3}{x - 1} \cdot \frac{x + 2}{x + 2} = \frac{(x + 3) \cdot (x + 2)}{(x - 1) \cdot (x + 2)} = \frac{x \cdot (x + 2) + 3 \cdot (x + 2)}{x \cdot (x + 2) - 1 \cdot (x + 2)} = \frac{x^{2} + 5 x + 6}{x^{2} + x + 4}$

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:

$\frac{x - 3}{x + 1} \cdot \frac{x - 4}{x - 4} = \frac{(x - 3) \cdot (x - 4)}{(x + 1) \cdot (x - 4)} = \frac{x \cdot (x - 4) - 3 \cdot (x - 4)}{x \cdot (x - 4) + 1 \cdot (x - 4)} = \frac{x^{2} - 7 x + 12}{x^{2} - 3 x - 4}$

Solution:

$\frac{x^{2} + 5 x + 6}{x^{2} + x + 4}$

$\frac{x^{2} - 7 x + 12}{x^{2} - 3 x - 4}$

Hide solution and development