1. ## Illegal Values

I must be growing senile in my old age. Else how can you explain I have simply forgotten how to do this problem

Find the illegal values of c in the multiplication statement c^2-3c-10 over c^2+5c-14 times c^2-c-2 over c^2-2c-15.
a.) c=7, c=3, c=-2, c=-5 b.) c=-7, c=-3, c=-2, and c=5 c.) c=-7, c=-3, c=2, and c=5 d.) c=-7 and c=-3

I reached as far as this ((c+2)(c+1)/(c+7)(c+3))

Now what? Something has to equal 0 except I dont know if its the numerator or the denominator.

Similarly
Find the illegal values of b in the fraction 2b^2+3b-10 over b^2-2b-8.
a.) b=-5 and 2 b.) b=-2 and -4 c.) b=-2 and 4 d.) b=-5, -2, 2 and 4

I must be growing senile in my old age. Else how can you explain I have simply forgotten how to do this problem

Find the illegal values of c in the multiplication statement c^2-3c-10 over c^2+5c-14 times c^2-c-2 over c^2-2c-15.
a.) c=7, c=3, c=-2, c=-5 b.) c=-7, c=-3, c=-2, and c=5 c.) c=-7, c=-3, c=2, and c=5 d.) c=-7 and c=-3

I reached as far as this ((c+2)(c+1)/(c+7)(c+3))

Now what? Something has to equal 0 except I dont know if its the numerator or the denominator.

Similarly
Find the illegal values of b in the fraction 2b^2+3b-10 over b^2-2b-8.
a.) b=-5 and 2 b.) b=-2 and -4 c.) b=-2 and 4 d.) b=-5, -2, 2 and 4

One of the rules of this forum is not to discuss something which is illegal

Illegal values over here is where the function is undefined.
Thus,
$\frac{c^2-3c-10}{c^2+5c-14}$ factors as,
$\frac{(c-5)(c+2)}{(c+7)(c-2)}$.
Notice that this is a fraction, and thus it is undefined where the denominator is zero. Thus, in this case,
$(c+7)(c-2)=0$ which gives the solutions,
$c=-7,2$.

3. Thank you.

I would like to ask whether it make a difference that the above term is also being multiplied by another expression since the question quite clearly says
c^2-3c-10 over c^2+5c-14 times c^2-c-2 over c^2-2c-15.

Thank you.

I would like to ask whether it make a difference that the above term is also being multiplied by another expression since the question quite clearly says
c^2-3c-10 over c^2+5c-14 times c^2-c-2 over c^2-2c-15.
It does not matter, you always look at the bottom.

5. Originally Posted by ThePerfectHacker
It does not matter, you always look at the bottom.
He is trying to tell you the question asked for the illegal values of $c$ in:

$
\frac{c^2-3c-10}{c^2+5c-14} \times \frac{c^2-c-2} {c^2-2c-15}
$

RonL

6. So basically the final answer should be that the illegal values of C are -7 and -3
Right?

So basically the final answer should be that the illegal values of C are -7 and -3
Right?
Yes, if you are allowed to cancel the factors of the terms at the top and
bottom which are equal, but strictly the illegal values are:

c.) c=-7, c=-3, c=2, and c=5

The ratio is indeterminate for c=2 and c=5, but the limits exist at these points.

RonL

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### illegal value

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