Answer:
The factored form is :
![6n(n-1)(n^(2)-3n-3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/xpapuzpc8nw77bjld5nwmmjmqrpbscg4nc.png)
Explanation:
The given expression is :
![6n^(4) -24n^(3) +18n](https://img.qammunity.org/2020/formulas/mathematics/middle-school/nmd7rfsdiumwhsglbf1iv2176wyiladyu5.png)
Taking out 6n out as 6n is common for all, we get
![6n(n^(3) -4n^(2) +3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/iubki3ch9u34yz8vykzaorajtjsxkjei0r.png)
Now lets factor
by hit and trial method.
Putting n=1
Now, by hit and trial method, we put n=1,
p(n)=
![1^(3) -4(1)^(2) +3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/bti6qomaj0b09bbq8wgawwxdmmwkq1blhs.png)
=> p(1) =
![1-4+3=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/c0vnm621l55cz4lzvznlkbx2b6rjd9umal.png)
So, (n-1) is a factor.
Now, dividing
by n-1 we get
![n^(2)-3n-3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/o1cpbillefnn6h58sqien2mrrkglthjjm5.png)
Therefore, the factored form is =
![6n(n-1)(n^(2)-3n-3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/xpapuzpc8nw77bjld5nwmmjmqrpbscg4nc.png)