Question

find the zeroes of the cubic polynomial x³-12x²+39x-28,if the zeroes ar in AP

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