How do you simplify and restricted value of #(x^3 - 2x + 3)/( x^2 + 12x + 32)#?

1 Answer
May 19, 2017

Answer:

#frac(x^(3) - 2 x + 3)((x + 6)^(2))#; #x ne - 6#

Explanation:

We have: #frac(x^(3) - 2 x + 3)(x^(2) + 12 x + 32)#

Let's factorise the denominator using the "middle-term break":

#= frac(x^(3) - 2 x + 3)(x^(2) + 6 x + 6 x + 32)#

#= frac(x^(3) - 2 x + 3)(x(x + 6) + 6(x + 6))#

#= frac(x^(3) - 2 x + 3)((x + 6)(x + 6))#

#= frac(x^(3) - 2 x + 3)((x + 6)^(2))#

Now let's determine the restricted values of #x#.

The denominator of the fraction can never equal to zero:

#Rightarrow (x + 6)^(2) ne 0#

#Rightarrow x + 6 ne 0#

#Rightarrow x ne - 6#

#therefore frac(x^(3) - 2 x + 3)(x^(2) + 12 x + 32) = frac(x^(3) - 2 x + 3)((x + 6)^(2))#; #x ne - 6#