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

when factorizing answer is ,

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

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

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

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....

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

okay thank you

WELCOME....

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

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