How do you simplify #(x^2+4x-32)/(x+8)# and what are the ecluded values fot he variables?

Question

Avery Alexander · Accepted Answer

#x - 4#; #x ne - 8#

We have: #frac(x^(2) + 4 x - 32)(x + 8)#

Let's begin by factorising the numerator using the "middle-term break":

#= frac(x^(2) + 8 x - 4 x - 32)(x + 8)#

#= frac(x (x + 8) - 4 (x + 8))(x + 8)#

#= frac((x + 8)(x - 4))(x + 8)#

We can now cancel the #x + 8# term:

#= x - 4#

Now, let's determine the excluded value of the variable #x#.

The denominator of a fraction can never equal to zero.

Let's set the denominator of the original fraction equal to zero and solve for #x#:

#Rightarrow x + 8 = 0#

#therefore x = - 8#

So the excluded value of #x# is #- 8#.

Therefore, #frac(x^(2) + 4 x - 32)(x + 8) = x - 4#; #x ne - 8#.

Eli Assouline · Answer

undefined To simplify the expression (x^2+4x-32)/(x+8), we can factor the numerator and then cancel out any common factors with the denominator. The numerator can be factored as (x+8)(x-4). Therefore, the simplified expression is (x-4).

The excluded values for the variable x are the values that make the denominator equal to zero, since division by zero is undefined. In this case, the excluded value is x = -8, as it would make the denominator (x+8) equal to zero.