Verify that 3,-2,1 are the zeros …

p(x) = x³ - 2x² - 5x + 6
Put x = 3, we get
p(3) = (3)³ - 2(3)² - 5(3) + 6
= 27 - 18 - 15 + 6
p(3) = 0 ........... (i)
Put x = -2, we get
p(-2) = (-2)³ - 2(-2)² -5(-2) + 6
= -8 - 8 + 10 + 6
p(-2) = 0 ............ (ii)
Now put x = 1, we get
p(1) = (1)³ - 2(1)² -5(1) + 6
= 1 - 2 -5 + 6
p(1) = 0 ............. (iii)
From (i), (ii) and (iii)
3, -2, 1 are the zeros of p(x) = x³ -2x² - 5x + 6
{tex}\therefore \space \alpha = 3 , \space \beta = -2 \space and \space \gamma = 1{/tex}
Comparing the given polynomial with
p(x) = ax³ + bx² + cx + d
We get, a = 1, b = -2, c = -5 and d = 6
Now, {tex}(\alpha + \beta + \gamma){/tex}= (3 -2 + 1) = 2 = {tex}-\frac{b}{a}{/tex}
{tex}(\alpha\beta + \beta\gamma + \gamma \alpha){/tex} = [3(-2) + (-2)(1) + (1)(3)]
= (-6 - 2 + 3) = -5 = {tex}\frac{c}{a}{/tex}
and {tex}\alpha\beta\gamma{/tex} = [3(-2)(1)] = -6 = {tex}-\frac{d}{a}{/tex}