ANSWER
[tex]\begin{gathered} \text{ Rational expression in lowest terms: }y-3 \\ \\ \text{ Variable restrictions for the original expression: }y\ne2,-3 \end{gathered}[/tex]
EXPLANATION
We want to reduce the rational expression to the lowest terms:
[tex]\frac{y^3-2y^2-9y+18}{y^2+y-6}[/tex]
First, let us factor the denominator of the expression:
[tex]\begin{gathered} y^2+y-6 \\ \\ y^2+3y-2y-6 \\ \\ y(y+3)-2(y+3) \\ \\ (y-2)(y+3) \end{gathered}[/tex]
Now, we can test if the factors in the denominator are also the factors in the numerator.
To do this for (y - 2), substitute y = 2 in the numerator. If it is equal to 0, then, it is a factor:
[tex]\begin{gathered} (2)^3-2(2)^2-9(2)+18 \\ \\ 8-8-18+18 \\ \\ 0 \end{gathered}[/tex]
Since it is equal to 0, (y - 2) is a factor. Now, let us divide the numerator by (y -2):
We have simplified the numerator and now, we can factorize by the difference of two squares:
[tex]\begin{gathered} y^2-9 \\ \\ y^2-3^2 \\ \\ (y-3)(y+3) \end{gathered}[/tex]
Therefore, the simplified expression is:
[tex]\frac{(y-2)(y-3)(y+3)}{(y-2)(y+3)}[/tex]
Simplify further by dividing common terms. The expression becomes:
[tex]y-3[/tex]
That is the rational expression in the lowest terms.
To find the variable restrictions, set the denominator of the original expression to 0 and solve for y:
[tex]\begin{gathered} y^2+y-6=0 \\ \\ y^2+3y-2y-6=0 \\ \\ y(y+3)-2(y+3)=0 \\ \\ (y-2)(y+3)=0 \\ \\ y=2,\text{ }y=-3 \end{gathered}[/tex]
Those are the variable restrictions for the original expression.
