Solution

In[67]:=

(*1*)y = (6 + 11x + 6x * x + x * x * x)/(-x + x * x * x)

Out[67]=

(6 + 11 x + 6 x^2 + x^3)/(-x + x^3)

In[68]:=

(*2*)Expand[y]

Out[68]=

6/(-x + x^3) + (11 x)/(-x + x^3) + (6 x^2)/(-x + x^3) + x^3/(-x + x^3)

In[69]:=

Factor[y]

Out[69]=

((2 + x) (3 + x))/((-1 + x) x)

In[70]:=

Together[y]

Out[70]=

(6 + 5 x + x^2)/((-1 + x) x)

In[71]:=

Apart[y]

Out[71]=

1 + 12/(-1 + x) - 6/x

In[72]:=

Cancel[y]

Out[72]=

(6 + 5 x + x^2)/((-1 + x) x)

In[73]:=

Simplify[y]

Out[73]=

(6 + 5 x + x^2)/((-1 + x) x)

In[74]:=

Numerator[y]

Out[74]=

6 + 11 x + 6 x^2 + x^3

In[75]:=

Denominator[y]

Out[75]=

-x + x^3

In[76]:=

(*3*)Solve[Numerator[y] 0, x]

Out[76]=

{{x -3}, {x -2}, {x -1}}

In[77]:=

(*4*)y0/.%

1 Power :: infy : Infinite expression - encountered. More… 0

∞ :: indet : Indeterminate expression 0 ComplexInfinity encountered. More…

Out[77]=

{True, True, Indeterminate0}

The solution x=-1 is a zero of the denominator of y. This can be seen from the various simplified versions of y, where common factors have been cancelled out. Since x=-1 is a zero of the denominator, y is not defined there. We could just drop the third solution alltogether. Another way to proceed is to solve instead for the zeros of the numerator of one of the simplified versions of y, for example Factor[y]. These are of course trivial.

In[78]:=

Solve[Numerator[Factor[y]] 0, x]

Out[78]=

{{x -3}, {x -2}}

In[79]:=

Clear[y]

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