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We want to divide the given rational expressions. $x_{2}+2x+1x_{2} ÷x_{2}−13x $
We will begin by factoring the numerators and denominators of each expression — if possible — starting with the first. Since the numerator is already factored, we will factor only the denominator.
Now, let's factor the second expression.
To divide the expressions, we will multiply the first expression by the reciprocal of the second expression.
$(x+1)(x+1)x_{2} ÷(x+1)(x−1)3x ⇕(x+1)(x+1)x_{2} ⋅3x(x+1)(x−1) $
Finally, we can cancel out any common factors.
To identify the restrictions on the variable, we need to find any values of $x$ that would cause the denominator of the simplified expression, or **any other denominator used**, to be $0.$ For simplicity, we will use their factored forms.

$x_{2}+2x+1x_{2} $

$(x+1)(x+1)x_{2} $

$x_{2}−13x $

WritePowWrite as a power

$x_{2}−1_{2}3x $

FacDiffSquares$a_{2}−b_{2}=(a+b)(a−b)$

$(x+1)(x−1)3x $

$(x+1)(x+1)x_{2} ⋅3x(x+1)(x−1) $

SplitIntoFactorsSplit into factors

$(x+1)(x+1)x⋅x ⋅3⋅x(x+1)(x−1) $

CancelCommonFacCancel out common factors

$(x+1) (x+1)x⋅x ⋅3⋅x (x−1)(x+1) $

SimpQuotSimplify quotient

$3(x+1)x(x−1) $

Denominator | Restrictions on the denominator | Restrictions on the variable |
---|---|---|

$(x+1)(x+1)$ | $x+1 =0$ | $x =-1$ |

$(x+1)(x−1)$ | $x+1 =0$ and $x−1 =0$ | $x =-1$ and $x =1$ |

$3x$ | $3x =0$ | $x =0$ |

$x+1$ | $x+1 =0$ | $x =-1$ |

We found three restrictions on the variable. $x =-1,x =0,andx =1, $