Question

find the zeroes of the polynomial f(x)=x³-12x²+39x-28, if the zeroes are in A.P.

Answer 1

f(x)=x3−12x2+39x−28f(x)=x3−12x2+39x−28

Let a-d, a, a+d be the zeroes of the Polynominal

Sum of Zeroes=−(−121)=12Sum of Zeroes=−(−121)=12

(i.e) a-d + a + a+d = 12

⟹3a=12ora=4⟹3a=12ora=4

Product of Zeroes=−(−281)=28Product of Zeroes=−(−281)=28

(a-d) (a) (a+d) = 28

a(a2−d2)=28But a=4a(a2−d2)=28But a=4

(a2−d2)=7(a2−d2)=7

(i.e)16−d2=7⟹d2=9(i.e)16−d2=7⟹d2=9

Therefore d=±3Therefore d=±3

Hence the roots are:

(3 - 4) (4) (3 + 4) or (-3 - 4) (4) (-3 + 4)

(i.e) -1, 4, 7 or -7, 4, 1