Respuesta :
Let [tex]f(x) 6n^4-24n^3+18n[/tex]
First thing I notice is that all those terms are multiples of 6 and n so we can pull those straight out,
[tex]f(x) = 6n(n^3-4n^2+3)[/tex]
Now the trickier part, notice if we let n=1 then [tex](n^3-4n^2+3) = 0[/tex]. By the factor-remainder theorem this means [tex](n-1)[/tex] is a factor of [tex](n^3-4n^2+3)[/tex].
Now perform polynomial division, dividing [tex](n-1)[/tex] into [tex]n^3-4n^2+3[/tex]. You are left with the quotient [tex](n^2-3n-3)[/tex] hence
[tex]f(x)=6n(n-1)(n^2-3n-3)[/tex]
First thing I notice is that all those terms are multiples of 6 and n so we can pull those straight out,
[tex]f(x) = 6n(n^3-4n^2+3)[/tex]
Now the trickier part, notice if we let n=1 then [tex](n^3-4n^2+3) = 0[/tex]. By the factor-remainder theorem this means [tex](n-1)[/tex] is a factor of [tex](n^3-4n^2+3)[/tex].
Now perform polynomial division, dividing [tex](n-1)[/tex] into [tex]n^3-4n^2+3[/tex]. You are left with the quotient [tex](n^2-3n-3)[/tex] hence
[tex]f(x)=6n(n-1)(n^2-3n-3)[/tex]
Answer: 6n(n^3-4n^2+3) of
Step-by-step explanation:
