\(\dfrac{b^2 + 10b + 24}{10b^2+40b} = \\ \dfrac{ b^2 + 10 b + 24}{10b(b+4)} = \\ \dfrac{(b+6)(b+4)}{10b(b+4)}=\\ \dfrac{b+6}{10b}\)
The only excluded value is
\(b=0\)
The numerator is a polynomial......so any real number b will "work"
We only have to worry about the denominator being 0
To find which values make this so we have
10b^2 + 40b = 0 factor
10b ( b + 4) = 0
So either 10b = 0 and b = 0 or b + 4 = 0 and b = -4
So...the excluded values are 0 and - 4
The simplified rational function is just as Rom found....