(i) 2x³+x²-5x+2; -1/2, 1, -2

Solution:

Given, p(x) = 2x³+x²-5x+2

And zeroes for p(x) are = 1/2, 1, -2

∴ p(1/2) = 2(1/2)³+(1/2)²-5(1/2)+2 = (1/4)+(1/4)-(5/2)+2 = 0

p(1) = 2(1)³+(1)²-5(1)+2 = 0

p(-2) = 2(-2)³+(-2)²-5(-2)+2 = 0

Hence, proved 1/2, 1, -2 are the zeroes of 2x³+x²-5x+2.

Now, comparing the given polynomial with general expression, we get;

∴ ax³+bx²+cx+d = 2x³+x²-5x+2

a=2, b=1, c= -5 and d = 2

As we know, if α, β, γ are the zeroes of the cubic polynomial ax³+bx²+cx+d , then;

α +β+γ = –b/a

αβ+βγ+γα = c/a

α βγ = – d/a.

Therefore, putting the values of zeroes of the polynomial,

α+β+γ = ½+1+(-2) = -1/2 = –b/a

αβ+βγ+γα = (1/2×1)+(1 ×-2)+(-2×1/2) = -5/2 = c/a

α β γ = ½×1×(-2) = -2/2 = -d/a

Hence, the relationship between the zeroes and the coefficients are satisfied.

(ii) x³-4x²+5x-2 ;2, 1, 1

Solution:

Given, p(x) = x³-4x²+5x-2

And zeroes for p(x) are 2,1,1.

∴ p(2)= 2³-4(2)²+5(2)-2 = 0

p(1) = 1³-(4×1² )+(5×1)-2 = 0

Hence proved, 2, 1, 1 are the zeroes of x³-4x²+5x-2

Now, comparing the given polynomial with general expression, we get;

∴ ax³+bx²+cx+d = x³-4x²+5x-2

a = 1, b = -4, c = 5 and d = -2

As we know, if α, β, γ are the zeroes of the cubic polynomial ax³+bx²+cx+d , then;

α + β + γ = –b/a

αβ + βγ + γα = c/a

α β γ = – d/a.

Therefore, putting the values of zeroes of the polynomial,

α +β+γ = 2+1+1 = 4 = -(-4)/1 = –b/a

αβ+βγ+γα = 2×1+1×1+1×2 = 5 = 5/1= c/a

αβγ = 2×1×1 = 2 = -(-2)/1 = -d/a

Hence, the relationship between the zeroes and the coefficients are satisfied.