# Illegal Values

• Mar 1st 2006, 09:57 AM
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

• Mar 1st 2006, 10:10 AM
ThePerfectHacker
Quote:

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 :D

Illegal values over here is where the function is undefined.
Thus,
$\displaystyle \frac{c^2-3c-10}{c^2+5c-14}$ factors as,
$\displaystyle \frac{(c-5)(c+2)}{(c+7)(c-2)}$.
Notice that this is a fraction, and thus it is undefined where the denominator is zero. Thus, in this case,
$\displaystyle (c+7)(c-2)=0$ which gives the solutions,
$\displaystyle c=-7,2$.
• Mar 1st 2006, 04:27 PM
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.
• Mar 1st 2006, 06:09 PM
ThePerfectHacker
Quote:

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.
• Mar 1st 2006, 09:11 PM
CaptainBlack
Quote:

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 $\displaystyle c$ in:

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

RonL
• Mar 2nd 2006, 05:37 AM
So basically the final answer should be that the illegal values of C are -7 and -3
Right?
• Mar 2nd 2006, 08:52 AM
CaptainBlack
Quote: