Answer
$a.$
$\displaystyle \frac{x-6}{x^{2}+x-6}=\frac{A}{x+3}+\frac{B}{x-2}$
$b.$
$\displaystyle \frac{x^{2}}{x^{2}+x+6}=1-\frac{x+6}{x^{2}+x+6}$
The denominator can't be factored.
Work Step by Step
a.
Factor the denominator:
$x^{2}+x-6=(x+3)(x-2).$
No repeated factors.
$\displaystyle \frac{x-6}{x^{2}+x-6}=\frac{A}{x+3}+\frac{B}{x-2}$
$b.$
The numerator has the same degree as the denominator. Either use polynomial division or
$\displaystyle \frac{(x^{2}+x+6)-x-6}{x^{2}+x+6}=1-\frac{x+6}{x^{2}+x+6}$
The denominator is irreducible (has no real zeros) because
$b^{2}-4ac=1-24$ is negative.
