Question

Find all other zeroes of the polynomial x^4+x^3-9x^2-3x+18 if two of its zeroes are root 3 and root -3.

Answer 1

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Solution is given in above attachment ..
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Answer 2

Heya !!!




( root 3 ) and (- root 3) are the two zeroes of the given polynomial.





( X - root 3 ) ( X + root 3 ) are also factor of polynomial P(X).


Therefore,


( X - root 3 ) ( X + root 3) = (X² - 3)





G(X) = X²-3




P(X) = X⁴ + X³ - 9X² - 3X+ 18




On dividing P(X) by G(X) we get,





X² - 3 ) X⁴ + X³ - 9X² - 3X + 18 ( X² + X -6


*********X⁴ ******-3X²


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******0****+X³ - 6X² - 3X + 18



***********X³ *********-3X



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*******0****-6X² ****0*****+18


***********-6X² ************+18

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****************0********************




We get,




Remainder = 0



And,


Quotient = X² + X - 6



After factorise the quotient we will get two other zeroes of the given polynomial.






=> X²+X -6



=> X² + 3X - 2X -6




=> X ( X + 3) - 2 ( X +3)






=> (X + 3) ( X -2) = 0






=> (X + 3) = 0 OR (X -2) = 0







=> X = -3 OR X = 2



Hence,



-3 , root 3 , 2 and - root 3 are four zeroes of the polynomial X⁴+X³-9X² -3X + 18.




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