+37157 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) )
+130071 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
+33652 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!
