Using Conjugate Roots Theorem, factor the polynomial function completely

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

Apr 11th, 2015

when factorizing answer is ,

x^6−73x^4−657x^2+729=(x−1)(x+1)(x−9)(x^2+9)(x+9).......................



Apr 11th, 2015

the roots of the polynomial are,

x1=1

x2=−1

x3=9

x4=−9

x5=3i

x6=−3i        

                        3i and -3i are complex roots , all the others are real roots

 


Apr 11th, 2015

This is the exact same answer I come up with and the computer is saying its not completely factored.

Apr 11th, 2015

x^6−73x^4−657x^2+729=(x−1)(x+1)(x−9)(x^2+9)(x+9)

            this is equal to,

                                            =(x−1)(x+1)(x−9) (x+3i)(x-3i)(x+9)...................

this is factorized completely......

here there are 6 roots because degree is 6 

here (x+3i)(x-3i) forms the conjugate roots....

Apr 11th, 2015

(x−1)(x+1)(x−9) (x+3i)(x-3i)(x+9)...................................THIS IS THE ANSWER

Apr 11th, 2015

okay thank you

Apr 11th, 2015

WELCOME....

HOPE YOU UNDERSTOOD..............THANK YOU

Apr 11th, 2015

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Apr 11th, 2015
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