Question

rify that 5, –2 and 1

3

are the zeroes of the cubic polynomial p(x) = 3x3

– 10x2

– 27x + 10 and

verify the relation between its zeroes and coefficients.

Answer 1

we have
this is the correct answer

Answer 2

Answer:

p(x)= 3x^3-10x^2-27x+10.

p(5)= 3×(5)^3 -10(5)^2 -27×5 +10.

P(5)= 375- 250 -135 +10.P(5)=0.

P(-2)= 3(-2)^3 -10(-2)^2 -27(-2)+10.

P(-2)= -24 -40 +54 +10.

P(-2)= -64 +64 = 0.

P(1/3)= 3(1/3)^3 -10(1/3)^2 -27(1/3) +10.

P(1/3)= 1/9 -10/9- 9 +10 = 0.

alpha + beta + gamma = -b/a .

5 + (-2) + 1/3 = 10/3 or 3+1/3

3+1/3 = 3+1/3.

alpha beta + beta gamma + gamma alpha = c/a.

-10 + (-2/3) + 5/3 = -27/3

-32+5/3 = -27/3.

-27/3 = -27/3.

alpha beta gamma = -d/a .

5×(-2)×1/3 = -10/3

-10/3 = -10/3 .

LHS = RHS

verified .