Question 959556



a. find all the zeros

{{{f(x)=7x^3-62x^2+167x-102}}}, 
if {{{4-i }}}is a zero, then {{{4+i}}} is a zero

using zero product formula we have {{{(x-(4-i))(x-(4+i))}}}=>{{{x^2-8x+17}}}

now we need to use long division and divide {{{7x^3-62x^2+167x-102}}} by {{{x^2-8x+17}}}

--------------{{{7x-6}}}
{{{x^2-8x+17}}}|{{{7x^3-62x^2+167x-102 }}}
-------------{{{7x^3-56x^2+119x}}}
----------------.....{{{-6x^2+48x-102}}}
---------------......{{{-6x^2+48x-102}}}
---------------.............................{{{0}}} =>reminder
so, third zero is {{{7x-6}}}, and the product of all zeros is:

{{{f(x)=(7x-6)(x-(4-i))(x-(4+i))}}} or
{{{f(x)=(7x-6) (x^2-8x+17)}}}

b. factor {{{f(x)}}} as a product of linear factors

{{{f(x)=7x^3-62x^2+167x-102}}}...write {{{-62x^2}}} as {{{-56x^2-6x^2}}} and {{{167x}}} as {{{119x+48x}}} 
{{{f(x)=7x^3-56x^2-6x^2+119x+48x-102}}}......group
{{{f(x)=(7x^3-6x^2)-(56x^2-48x)+(119x-102)}}}

{{{f(x)=x^2(7x-6)-8x(7x-6)+17(7x-6)}}}

{{{f(x)=(7x-6)(x^2-8x+17)}}}

c. solve the equation {{{f(x)=0}}}

{{{(7x-6)(x^2-8x+17)=0}}} or like we have in a.{{{(x^2-8x+17)=(x-(4-i))(x-(4+i))}}}

{{{(7x-6)(x-(4-i))(x-(4+i))=0}}}

solutions:

{{{(7x-6)=0}}}=>{{{7x=6}}}=>{{{x=6/7}}}
and two already given zeros
{{{(4-i)}}}
{{{(4+i)}}}