Question

find the zeroes of the polynomial f(x) = `x^3`-`12x^2` +39x -28 , if the zeroes are in A.P

Answer 1

Let the roots of the given equation `x^3-12x^2+39x - 28 = 0` are,
`a-d, a and a+d` as they are in A.P.
Then, sum of roots,
`a-d+a+a+d = -(-12) `
`=> 3a = 12 => a= 4`
Product in pairs of the roots.
`a(a-d)+a(a+d)+(a-d)(a+d) = 39`
`=>a(a-d+a+d))+a^2-d^2 = 39`
`=>3a^2 - d^2 = 39`
`=>3(4)^2 -39 = d^2`
`=>d = +-3`
If `d = 3`, roots are `1,4 and 7`.
If `d = -3`, roots are `7,4 and 1`.