Answer:
The answer is "2, and 1/2 is not a part of polynomial".
Step-by-step explanation:
Given equation:
[tex]P(x) = 2x^3-3x^2-3x+2\\[/tex]
put the value 2, -1, and 1/2 in above equation we get:
p(2) = 0
p (-1)= 0
p (1/2) = 0
compare the above equation with: [tex]ax^3+bx^2+cx+d= 0[/tex]
a= 2
b= -3
c=-3
d= 2
[tex]i) \ \alpha \beta+ \beta\gamma+\gamma \alpha= \frac{c}{a}\\\\ ii) \ \alpha +\beta +\gamma= \frac{-b}{a}\\\\ iii) \ \alpha \beta \gamma= \frac{-d}{a}\\\\[/tex]
[tex]i) 2 \times -1 + -1 \times \frac{1}{2} + \frac{1}{2} \times 2 = \frac{-(3)}{2}\\\\-2- \frac{1}{2}+1= \frac{-3}{2}\\\\\frac{-4-1+2}{2} = \frac{-3}{2} \\\\\frac{-3}{2} =\frac{-3}{2}\\[/tex]
[tex]ii) \ 2 + (-1) + \frac{1}{2} = \frac{-(-3)}{2}\\\\2-1+\frac{1}{2} = \frac{3}{2}\\\\\frac{4-2+1}{2}= \frac{3}{2}\\\\\frac{3}{2}=\frac{3}{2}\\\\[/tex]
[tex]iii) 2 \times -1 \times \frac{1}{2} = \frac{-2}{2}\\\\ -1 =-1[/tex]
