Question

33. Find the condition that the zeroes of the polynomial f(x) = x^3+ 3ax'^2+ 3bx + c are in AP.​

Answer 1

Answer:

2a³ + c = 3ab

Step-by-step explanation:

Given---> Zeroes of the polynomial

f(x) = x³ + 3ax² + 3bx + c are in AP

To find ---> Condition that zeroes of given polynomial is in AP.

Solution--->

f (x ) =x³ + 3ax² + 3bx + c

Let roots of given polynomial be α ,β and γ

Roots are in AP ATQ so

α + γ = 2β

α + β + γ = - coefficient of x² / coefficient of x³

α + β + γ = - 3 a / 1

(α + γ )+ β = - 3a

Putting α + γ = 2β in it

2 β + β = - 3a

3β = - 3a

β = - a

Now

α β γ = -constant terms/coefficient ofx³

= - c / 1

α γ ( β ) = - c

Putting β = -a in it

α γ ( - a ) = - c

α γ = c / a

Now

αβ + βγ + γα

= coefficient of x / Coefficient of x³

= 3b / 1

αβ + βγ + γα = 3b

β ( α + γ ) + γα = 3b

Putting β = -a and γα = c / a

(-a ) ( 2β ) + c/a = 3b

Putting β = -a in it

( -a ) ( - 2a ) + c / a = 3b

2 a² + c / a = 3b

2 a³ + c / a = 3 b

2 a³ + c = 3ab