Answer:
The correct option is B
Step-by-step explanation:
we have to find the illegal values of b in the fraction
\(\frac{2b^2+3b-10}{b^2-2b-8}\)
The values of b that makes the denominator 0 which are illegal values of b in the fraction
\(\frac{2b^2+3b-10}{b^2-2b-8}\)
\(b^2-2b-8=0\)
solve factor
\(b^2-2b-8=0\)
By middle term splitting method
\(b^2-4b+2b-8=0\)
\(b(b-4)+2(b-4)\)
\((b-4)(b+2)\)
b=4 and -2
The correct option is B
