Respuesta :

Answer: Its factored form will be

[tex]6n(n-1)(n^2-3n-3)[/tex]

Step-by-step explanation:

Since we have given that

[tex]6n^4-24n^3+18n[/tex]

So, we need to factor ,

[tex]6n(n^3-4n^2+3)[/tex]

Now,

[tex]\text{Let }p(n)=n^3-4n^2+3[/tex]

Now, by hit and trial method, we put n=1,

[tex]p(1)=1^3-4\times1^2+3=1-4+3=1-1=0[/tex]

[tex]n=1\\n-1=0[/tex]

So, n-1 is also factor of p(n).

Now,

[tex]\frac{n^3-4n^2+3}{n-1}=n^2-3n-3[/tex]

So, its factored form will be

[tex]6n(n-1)(n^2-3n-3)[/tex]

Answer:

6n(n^3-4n^3+3) or C on e2020

Step-by-step explanation:

The GCF is 6n. divide all the terms by 6n.