ANSWER
[tex]\frac{n^4-10n^2+24}{n^4-9n^2+18}=\frac{(n-2)(n+2)}{(n-\sqrt{3})(n+\sqrt{3})}[/tex]
For
[tex]n\ne \pm\sqrt{3}[/tex]
EXPLANATION
We have [tex]\frac{n^4-10n^2+24}{n^4-9n^2+18}[/tex]
This is very easy to simplify. We shall look at the two expressions from a quadratic trinomial perspective.
We rewrite the rational expression to obtain;
[tex]\frac{n^4-10n^2+24}{n^4-9n^2+18}=\frac{(n^2)^2-10(n^2)+24}{(n^2)^2-9(n^2)+18}[/tex]
We can now see that both the numerator and denominator are quadratic trinomials in [tex]n^2[/tex].
We split the middle terms as follows;
[tex]\frac{n^4-10n^2+24}{n^4-9n^2+18}=\frac{(n^2)^2-6n^2-4n^2+24}{(n^2)^2-6n^2-3n^2+18}[/tex]
[tex]\frac{n^4-10n^2+24}{n^4-9n^2+18}=\frac{n^2(n^2-6)-4(n^2-6)}{n^2(n^2-6)-3(n^2-6)}[/tex]
We factor further to obtain;
[tex]\frac{n^4-10n^2+24}{n^4-9n^2+18}=\frac{(n^2-6)(n^2-4)}{(n^2-6)(n^2-3)}[/tex]
We now cancel out common factors to get;
[tex]\frac{n^4-10n^2+24}{n^4-9n^2+18}=\frac{(n^2-4)}{(n^2-3)}[/tex]
[tex]\frac{n^4-10n^2+24}{n^4-9n^2+18}=\frac{(n-2)(n+2)}{(n-\sqrt{3})(n+\sqrt{3})}[/tex]
For
[tex]n\ne \pm\sqrt{3}[/tex]
1. Factor expression [tex]n^4-10n^2+24:[/tex]
[tex]n^4-10n^2+24=(n^2-4)(n^2-6)=(n-2)(n+2)(n-\sqrt{6})(n+\sqrt{6}).[/tex]
2. Factor expression [tex]n^4-9n^2+18:[/tex]
[tex]n^4-9n^2+18=(n^2-3)(n^2-6)=(n-\sqrt{3})(n+\sqrt{3})(n-\sqrt{6})(n+\sqrt{6}).[/tex]
Note that
[tex]n\neq \sqrt{3},\\n\neq -\sqrt{3},\\n\neq \sqrt{6},\\n\neq -\sqrt{6},[/tex]
because this expression is placed in the denominator.
3. Now
[tex]\dfrac{n^4-10n^2+24}{n^4-9n^2+18}=\dfrac{(n-2)(n+2)(n-\sqrt{6})(n+\sqrt{6})}{(n-\sqrt{3})(n+\sqrt{3})(n-\sqrt{6})(n+\sqrt{6})}=\\ \\=\dfrac{(n-2)(n+2)}{(n-\sqrt{3})(n+\sqrt{3})}=\dfrac{n^2-4}{n^2-3}.[/tex]
