Determining if ( x^{3} ) is a Factor of the Polynomial ( 2x^{3} - 9x^{2} 1 - 3 )

To determine if ( x^{3} ) is a factor of the polynomial ( 2x^{3} - 9x^{2} 1 - 3 ), we can use the Factor Theorem and polynomial division. According to the Factor Theorem, ( x - c ) is a factor of a polynomial ( P(x) ) if ( P(c) 0 ).

Using the Factor Theorem

In this case, we have ( c 3 ) so we will evaluate ( P(3) ).

Let's calculate ( P(3) ) for the polynomial ( P(x) 2x^{3} - 9x^{2} 1 - 3 ):

[ P(x) 2x^{3} - 9x^{2} 1 - 3 ]

Substituting ( x 3 ):

[ P(3) 2(3)^{3} - 9(3)^{2} 10(3) - 3 2(27) - 9(9) 30 - 3 ]

[ 54 - 81 30 - 3 54 - 84 3 0 ]

Since ( P(3) 0 ), it follows that ( x - 3 ) is a factor of ( 2x^{3} - 9x^{2} 1 - 3 ).

Using Polynomial Division

Alternatively, we can use polynomial division to confirm that ( x - 3 ) is a factor. We divide ( 2x^{3} - 9x^{2} 1 - 3 ) by ( x - 3 ):

 2x^2   3x   1 
___________
x - 3 | 2x^3 - 9x^2   1 - 3
      2x^3 - 6x^2 
      __________
            -3x^2   1
            -3x^2   9x
            _________
                   x - 3
                   x - 3
                   _____
                        0

The result of the division is ( 2x^2 3x 1 ), which means:

[ 2x^{3} - 9x^{2} 1 - 3 (x - 3)(2x^2 3x 1) ]

This confirms that ( x - 3 ) is indeed a factor of the polynomial.

Conclusion

Thus, ( x - 3 ) (or equivalently ( x^{3} )) is a factor of the polynomial ( 2x^{3} - 9x^{2} 1 - 3 ) because substituting ( 3 ) into the polynomial yields zero, and the division confirms the factorization.

Related Keywords: polynomial factor, factor theorem, polynomial division