Question 1114960


given: {{{n=3}}}, {{{-5}}} and {{{4+3i}}} are zeros; {{{f(2)=273}}} 

if {{{n=3}}}, we have a {{{3rd}}} degree function

Since {{{4+3i}}} is a root of f(x) and f(x) has real coefficients, {{{ 4-3i}}} is also a root.  

use zero product theorem:

{{{f(x) = a(x-(-5))(x-(4+3i))(x-(4-3i ))}}}

{{{f(x) = a(x+5)(x-4-3i)(x-4+3i )}}}....multiply all that and you  will get

{{{f(x) = a(x^3 - 3x^2 - 15x + 125)}}} ....since {{{f(2)=273}}}, use it to find {{{a}}}


{{{273 = a(2^3 - 3*2^2 - 15*2 + 125)}}}

{{{273 = 91a}}}

{{{a=273/91}}}

{{{a=3}}}

so, your function is:

{{{f(x) = 3(x^3 - 3x^2 - 15x + 125)}}}
 
{{{f(x) = 3x^3 - 9x^2 - 45x + 375}}}-> your function