Using the Conjugate Roots Theorem, factor the polynomial function completely

Algebra
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f(x)=x^6-73x^4-657x^2+729

Apr 11th, 2015

let x^2=A

f(x)=a^3-73a^2-657a+729

from observation, a=1 is a root

therefore

x^2=1

x= +/- 1

on further analysis, x=9 and x=3i also satisfy the equation.

Using the Conjugate theorem, 

x=-3i is also a root of the equation. So is x=-9

Hence using the above Values,

the factored polynomial is

(x−1)(x+1)(x−9)(x+9)(x+3i)(x-3i)

that is

(x−1)(x+1)(x−9)(x+9)(x^2+9)


Apr 11th, 2015

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