Given:
The zeros of degree 3 polynomial are -2, 3 , 6.
The coefficient of x² is -14.
Let the degree 3 polynomial be,
[tex]\begin{gathered} p(x)=(x-x_1)(x-x_2)(x-x_3) \\ =(x-(-2))(x-3)(x-6) \\ =\mleft(x+2\mright)\mleft(x-3\mright)\mleft(x-6\mright) \\ =\mleft(x^2-x-6\mright)\mleft(x-6\mright) \\ =x^3-x^2-6x-6x^2+6x+36 \\ =x^3-7x^2+36 \end{gathered}[/tex]But given that, coefficient of x² is -14 so, multiply the above polynomial by 2.
[tex]\begin{gathered} p(x)=x^3-7x^2+36 \\ 2p(x)=2(x^3-7x^2+36) \\ =2x^3-14x^2+72 \end{gathered}[/tex]Answer: The polynomial is,
[tex]p(x)=2x^3-14x^2+72[/tex]