Question

Find the zeroes of the polynomial $f\left(x\right)=x^3-12x^2+39x-28$, if it is given that zeroes are in $AP$.

Answer 1

$f\left(x\right)=x^3-12x^2+19x-28$

Zeroes are in AP.

Let the zeroes be of the form

a-d,a,a+d

sum of roots = 12

3a=12

a = 4

product of roots $=-\frac{d}{a}$

$a(a^2-d^2)=(-28)$

$a^3-a{d}^2=28$

$(4{)}^3-4d^2=28$

$64-4d^2=28$

$4d^2=36$

$d=\pm 3$

Zeroes are 1,4,7 or 7,4,1