SOLUTION: 7x^3-62x^2+167x-102, 4-i is a zero a, find all the zeros b,factor f(x)as a product of linear factors c,solve the equation f(x)=0

Algebra ->  Functions -> SOLUTION: 7x^3-62x^2+167x-102, 4-i is a zero a, find all the zeros b,factor f(x)as a product of linear factors c,solve the equation f(x)=0      Log On


   

Question 959556: 7x^3-62x^2+167x-102, 4-i is a zero
a, find all the zeros
b,factor f(x)as a product of linear factors
c,solve the equation f(x)=0
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!


a. find all the zeros
f%28x%29=7x%5E3-62x%5E2%2B167x-102,
if 4-i+is a zero, then 4%2Bi is a zero
using zero product formula we have %28x-%284-i%29%29%28x-%284%2Bi%29%29=>x%5E2-8x%2B17
now we need to use long division and divide 7x%5E3-62x%5E2%2B167x-102 by x%5E2-8x%2B17
--------------7x-6
x%5E2-8x%2B17|7x%5E3-62x%5E2%2B167x-102+
-------------7x%5E3-56x%5E2%2B119x
----------------.....-6x%5E2%2B48x-102
---------------......-6x%5E2%2B48x-102
---------------.............................0 =>reminder
so, third zero is 7x-6, and the product of all zeros is:
f%28x%29=%287x-6%29%28x-%284-i%29%29%28x-%284%2Bi%29%29 or
f%28x%29=%287x-6%29+%28x%5E2-8x%2B17%29
b. factor f%28x%29 as a product of linear factors
f%28x%29=7x%5E3-62x%5E2%2B167x-102...write -62x%5E2 as -56x%5E2-6x%5E2 and 167x as 119x%2B48x
f%28x%29=7x%5E3-56x%5E2-6x%5E2%2B119x%2B48x-102......group
f%28x%29=%287x%5E3-6x%5E2%29-%2856x%5E2-48x%29%2B%28119x-102%29
f%28x%29=x%5E2%287x-6%29-8x%287x-6%29%2B17%287x-6%29
f%28x%29=%287x-6%29%28x%5E2-8x%2B17%29
c. solve the equation f%28x%29=0
%287x-6%29%28x%5E2-8x%2B17%29=0 or like we have in a.%28x%5E2-8x%2B17%29=%28x-%284-i%29%29%28x-%284%2Bi%29%29
%287x-6%29%28x-%284-i%29%29%28x-%284%2Bi%29%29=0
solutions:
%287x-6%29=0=>7x=6=>x=6%2F7
and two already given zeros
%284-i%29
%284%2Bi%29