Question

Find the illegal values of b in the fraction (2b^2+36-10)/(b^2-26-8)

Answer 1

Correct Question is: Find the illegal values of b in the fraction (2b^2 + 3b - 10)/(b^2 - 2b - 8)

illegal values are those which are not a part of Domain. In this case these will be such values which will make the denominator of the fraction zero.

So setting the denominator equal to zero, we can find these values.

$b^{2} -2b-8=0 \\ \\ b^{2}-4b+2b-8=0 \\ \\ b(b-4)+2(b-4)=0 \\ \\ (b+2)(b-4)=0 \\ \\ b=-2, b=4$

Thus, b=-2 and b=4 are the illegal values for this expression