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Find the illegal values of b in the fraction

Find the illegal values of b in the fraction-example-1
User Tdavis
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1 Answer

7 votes

Answer:


b=4\text { or }b=-2

Step-by-step explanation:

We have been given a rational expression:
(2b^2+3b-10)/(b^2-2b-8) and we are asked to find the illegal values of b.

Illegal values of b will be those values, which will make our denominator 0.

To solve for illegal or excluded values of b, let us set our denominator equal to 0.


b^2-2b-8=0

We will factor out our equation by splitting the middle term.


b^2-4b+2b-8=0


b(b-4)+2(b-4)=0


(b-4)(b+2)=0


b-4=0\text { or }b+2=0


b=4\text { or }b=-2

Therefore,
b=4\text { or }b=-2 are illegal values of b in the given fraction.

User Maaw
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