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

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Answer 1 · 2017-05-19T21:52:50

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

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:

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

$Rightarrow x + 6 ne 0$

$Rightarrow x ne - 6$

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

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