#1**+1 **

First one p cannot equal - 4 (the denominator would be 0----> not allowed)

Factor the numerator (p-8)(p+4) then the (p+4) 'cancels' and you are left with

(p-8)

Corrected the p cannot equal part ....sorry. Answer is still the second choice.

See if you can do the second one...... (Hint: Denominator factors to (q-8)(q+3) )

ElectricPavlov
Feb 22, 2018

#2**+1 **

1. Notice, NSS...that p cannot = - 4 because the denominator would be = 0

Factoring p^2 - 4p - 32 we get ( p + 4) ( p - 8)

This would give us

[ (p + 4) ( p - 8) ] / (p + 4) the (p+4) 's cancel

So we're left with p - 8

So...the second answer is correct

2. [ q^2 + 11q + 24 ] / [ q^2 - 5q - 24 ] factor top/bottom

[ (q + 8) (q + 3) ] / [ (q - 8) (q + 3) ]

Notice that q cannot equal either 8 or -3 because these would make the original denominator = 0

Now.... the (q + 8) factors "cancel on top/botom and we're left with

(q +8 ) / ( q - 8)

So....the thiird answer is correct

CPhill
Feb 22, 2018

#3**+1 **

I don't agree that that the function \(\frac{p^2-4p-32}{p+4}\) is undefined at p = -4. Its value is just -12 there. Use L'Hopital's rule for example to find the limit.

Although the denominator goes to zero at p = -4, so too does the numerator - *at the same rate. * This negates the effect of the denominator going to zero.

You wouldn't say the function \(\frac{x}{x}\) is undefined at x = 0 simply because the denominator is zero there. The function is clearly 1 for *all *values of x.

If the numerator goes to zero as fast as, or faster than the denominator then the function will have a well defined value. It's only if the denominator goes to zero faster than the numerator that the function will have a singularity.

Of course, there might well be *numerical* problems in *evaluating *a function like \(\frac{p^2-4p-32}{p+4}\) at p = -4 if the numerical procedure first calculates numerator and denominator separately. However, the function is mathematically well-defined there!

Alan
Feb 23, 2018