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You are watching: Find the illegal values of c in the multiplication statement

I must be farming senile in my old age. Else how can you define I have simply forgotten how to execute this difficulty find the illegal values of c in the multiplication explain 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 also c=5 d.) c=-7 and c=-3I reached as much as this ((c+2)(c+1)/(c+7)(c+3))Now what? Something has to equal 0 except I dont know if that is the molecule or the denominator.Similarly uncover the illegal worths of b in the portion 2b^2+3b-10 end b^2-2b-8. A.) b=-5 and also 2 b.) b=-2 and -4 c.) b=-2 and 4 d.) b=-5, -2, 2 and 4 help Please
I have to be cultivation senile in my old age. Else how deserve to you define I have actually simply forgotten just how to perform this difficulty find the illegal values of c in the multiplication explain c^2-3c-10 over c^2+5c-14 time 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 also c=5 c.) c=-7, c=-3, c=2, and also c=5 d.) c=-7 and c=-3I reached as much as this ((c+2)(c+1)/(c+7)(c+3))Now what? Something has to equal 0 other than I dont recognize if the the molecule or the denominator.Similarly uncover the illegal worths of b in the portion 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 also 4 assist Please
One the the rule of this forum is no to comment on something i m sorry is illegal
Illegal values over below is whereby the role is undefined.Thus,(displaystyle fracc^2-3c-10c^2+5c-14) components as,(displaystyle frac(c-5)(c+2)(c+7)(c-2)).Notice the this is a fraction, and thus the is undefined where the denominator is zero. Thus, in this case,(displaystyle (c+7)(c-2)=0) which offers the solutions,(displaystyle c=-7,2).
Thank you.I would choose to ask even if it is it make a distinction that the over term is also being multiply by another expression since the question quite clearly says c^2-3c-10 over c^2+5c-14 time c^2-c-2 end c^2-2c-15.

Thank you.I would like to ask whether it do a distinction that the over term is likewise being multiplied by one more expression since the question quite clearly says c^2-3c-10 over c^2+5c-14 times c^2-c-2 end c^2-2c-15.

He is trying to tell you the inquiry asked because that the illegal values of (displaystyle c) in:(displaystyle fracc^2-3c-10c^2+5c-14 imes fracc^2-c-2 c^2-2c-15)RonL

Yes, if girlfriend are enabled to publication the components of the terms at the peak andbottom which are equal, but strictly the illegal worths are:c.) c=-7, c=-3, c=2, and also c=5 The ratio is indeterminate because that c=2 and also c=5, however the borders exist at these points.RonL
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