Question

find the zeroes of the ploynomial f(x)=x^3_12x^2+39x_28, if it is given that the zeroes are in A.p

Answer 1

Let zeroes be a-d, a and a+d having in AP.

By relation with coefficient,

a+d + a + a-d = 12
           3a = 12
 so, a = 4

 And (a-d)(a)(a+d) = 28
    so, (4-d)(4+d)(4) = 28
        so, 16 - d^2 = 7
            so, d^2 = 9
      so, d = +3 or -3

hence, zeroes are = 7, 4, and 1 
             

Answer 2

Given polynomial is f(x) = x^3 - 12x^2 + 39x - 28.

Given that the zeroes are in AP.

Let its zeroes be (a-d),a,(a+d).

The given equation is in the form of ax^3 + bx^2 + cx + d = 0

Where a  1, b = -12, c = 39, d = -28.

Now,

We know that Sum of the roots = -(b)/a

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

                   3a                             = 12

                      a = 4.


We know that product of the roots = -(d)/a

 (a - d) * a * (a + d) = -(-28)/1

Substitute a = 4, we get

(4 - d) * 4 * (4 + d) = 28

4(4 - d)(4 + d) = 28

(4 - d)(4 + d) = 28/4

(4 - d)(4 + d) = 7

16 - d^2 = 7

d^2 = 9

d = +3 (or) - 3.


Hence, the zeroes of the polynomial are (a - d),a,(a+d) = (4 - 3),4,(4+3)
      
                                                                                          =  1,4,7.


Hope this helps!