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

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Answer 1 · 2017-05-29T15:14:28

#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#.