Question

Simplify the rational expression. state any restrictions on the variable. n4-10n^2+24/n^4-9n^2+18

Answer 1

1. Factor expression
`n^4-10n^2+24:`

`n^4-10n^2+24=(n^2-4)(n^2-6)=(n-2)(n+2)(n-√(6))(n+√(6)).`

2. Factor expression
`n^4-9n^2+18:`

`n^4-9n^2+18=(n^2-3)(n^2-6)=(n-√(3))(n+√(3))(n-√(6))(n+√(6)).`

Note that

`n\\eq √(3),\\n\\eq -√(3),\\n\\eq √(6),\\n\\eq -√(6),`

because this expression is placed in the denominator.

3. Now

`(n^4-10n^2+24)/(n^4-9n^2+18)=((n-2)(n+2)(n-√(6))(n+√(6)))/((n-√(3))(n+√(3))(n-√(6))(n+√(6)))=\\ \\=((n-2)(n+2))/((n-√(3))(n+√(3)))=(n^2-4)/(n^2-3).`

Answer 2

`(n^4-10n^2+24)/(n^4-9n^2+18)=((n-2)(n+2))/((n-√(3))(n+√(3)))`

For

`n\\e \pm√(3)`

Step-by-step explanation

We have
`(n^4-10n^2+24)/(n^4-9n^2+18)`

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;

`(n^4-10n^2+24)/(n^4-9n^2+18)=((n^2)^2-10(n^2)+24)/((n^2)^2-9(n^2)+18)`

We can now see that both the numerator and denominator are quadratic trinomials in
`n^2`.

We split the middle terms as follows;

`(n^4-10n^2+24)/(n^4-9n^2+18)=((n^2)^2-6n^2-4n^2+24)/((n^2)^2-6n^2-3n^2+18)`

`(n^4-10n^2+24)/(n^4-9n^2+18)=(n^2(n^2-6)-4(n^2-6))/(n^2(n^2-6)-3(n^2-6))`

We factor further to obtain;

`(n^4-10n^2+24)/(n^4-9n^2+18)=((n^2-6)(n^2-4))/((n^2-6)(n^2-3))`

We now cancel out common factors to get;

`(n^4-10n^2+24)/(n^4-9n^2+18)=((n^2-4))/((n^2-3))`

`(n^4-10n^2+24)/(n^4-9n^2+18)=((n-2)(n+2))/((n-√(3))(n+√(3)))`

For

`n\\e \pm√(3)`