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# Help #1 ​

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Help #1

NotSoSmart  Feb 22, 2018
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#1
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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
edited by ElectricPavlov  Feb 22, 2018
edited by ElectricPavlov  Feb 22, 2018
#2
+86613
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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

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)

CPhill  Feb 22, 2018
edited by CPhill  Feb 22, 2018
#3
+26689
+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