Question

If the zeros of the polynomial f(x) = x3 - 3x2 - 6x + 8 are of the form a -b, a, a + b, find all the zeros​

Answer 1

Explanation:

=> p(x)= x³-3x²-6x+8

a= 1, b= -3,c= ‐6 &d=8

-b/a= -3/1 = -3

c/a= -6/1 = -6

-d/a= 8/1=8

therefore, Alpha +Beta+Gemma = -b/a= -3

Alpha ×beta + beta × Gemma +Gemma × Alpha = c/a= -6

Alpha ×Beta ×Gemma = -d/a = 8

so,all the zeros are -3,-6 &8