Question 174804
What are the "illegal" values of b in the fraction:
{{{(2b^2+3b-10)/(b^2-2b-8)}}}
The "illegal" (sometimes called "excluded") values of b are those values of b that would make the denominator equal zero. So, we set the denominator equal to zero and solve for b.
{{{b^2-2b-8 = 0}}} Solve by factoring.
{{{(b+2)(b-4) = 0}}} Apply the zero product rule.
{{{b+2 = 0}}} or {{{b-4 = 0}}}, so that...
{{{highlight(b = -2)}}} or {{{highlight(b = 4)}}} These are the "illegal" values of b.