Question 711193
1). We can factorize the numerator:

{{{p^2-4p-32}}}
 
then, we recognize the coefficients, the form of a quadratic equations is:

{{{ ax^2+bx+c }}}

so:

{{{ ap^2+bp+c }}}
{{{p^2-4p-32}}}

{{{ a=1}}}{{{ b=-4}}}{{{ c=-32}}}
Now, we can use quadratic equations:
*[invoke quadratic "p", 1, -4, -32 ]

roots are p=-4 and p = 8, so:

{{{p^2-4p-32 = (p+4)(p-8)}}}

then:

{{{ (p^2-4p-32)/(p+4) = (p+4)(p-8)/(p+4) = p-8}}} 

the restriction on this is: -4 because indetermine the expresion {{{(p^2-4p-32)/(p+4)}}} with zero in the denominator and this trend to infinite.

2) Again, We can factorize the numerator:

{{{q^2+ 11q +24}}}
{{{ aq^2+bq+c }}}

{{{ a=1}}}{{{ b=11}}}{{{ c=24}}}
*[invoke quadratic "q", 1, 11, 24 ]
roots are q = -8 and q = -3, so:

{{{q^2+ 11q +24 = (q+3)(q+8)}}}

too, We can factorize the denominator:
{{{q^2 -5q -24}}}
{{{ aq^2+bq+c }}}

{{{ a=1}}}{{{ b=-5}}}{{{ c=-24}}}
*[invoke quadratic "q", 1, -5, -24 ]
roots are q = 8 and q = -3, so:

{{{q^2+ 11q +24 = (q+3)(q-8)}}}

then:
{{{(q^2+ 11q +24)/ (q^2 -5q -24)=((q+3)(q+8))/((q+3)(q-8))=(q+8)/(q-8)}}}

the restriction on this is: 8 because indetermine the expresion {{{(q+8)/(q-8)}}} with zero in the denominator and this trend to infinite.

Did you understand me?