How do you simplify #(b^2+2b-24)/(2b^2-72)# and find any non permissible values?

Question

Avery Alexander · Accepted Answer

The expression simplifies to #(b - 4)/(2(b - 6))#, #b !=-6, 6#.

Factor everything. #=((b + 6)(b - 4))/(2(b^2 - 36))# #=((b + 6)(b - 4))/(2(b + 6)(b - 6))# #=(b - 4)/(2(b - 6))# For the non-permissible values, these will occur when the denominator equals #0#. #0 = 2b^2 - 72# #0 = 2(b^2 - 36)# #0 = (b + 6)(b - 6)# #b = -6 and 6# Hopefully this helps!

Julia Adams · Answer

undefined To simplify the expression (b^2+2b-24)/(2b^2-72), we can factor both the numerator and denominator.

The numerator can be factored as (b+6)(b-4), and the denominator can be factored as 2(b+6)(b-6).

Now, we can cancel out the common factors of (b+6) from both the numerator and denominator.

After canceling out the common factors, the simplified expression becomes (b-4)/(2b-12).

To find the non-permissible values, we need to identify the values of b that would make the denominator equal to zero.

Setting the denominator, 2b-12, equal to zero and solving for b, we get b = 6.

Therefore, the non-permissible value for b is 6, as it would result in division by zero.