Question

If the zeroes of the polynomial f(x) = x³ - 12x² + 39x + k are in A.P ,
Find the value of k

Answer 1

Here given, cubic polynomial .

hence,polynomial have three roots .
Let (a- d ), a and (a+ d) are the roots of given polynomial.
now,
x³ -12x² + 39x + K ,

sum of roots = - ( coefficient of x²)/(coefficient of x³)

(a - d)+ a + (a + d) = -(-12)/1 = 12

3a = 12

a = 4

again,
sum of products of two consequitive roots = ( coefficient of x)/(coefficient of x³)

(a - d)a + a(a + d) + (a + d)(a - d) = 39

a² -ad + a² + ad + a² -d² = 39

3a² - d² = 39

3(4)² - d² = 39

3 × 16 - d² = 39

d² = 9

d = ±3

hence,
roots are 1 , 4 , 7 or 7, 4 , 1

now,
products of all roots = - ( constant)/coefficient of x³

7 × 4 × 1 = - ( K)/1

K = -28

Answer 2

here is your answer