Find the condition that the zeros of the polynomial f(x) = x^3+3px^2+3px+r may be in a. p.

Step-by-step explanation:

note :

( a + b )^3 = a^3 +b^3 +3 a²b +3b²a + b^3...(*)

the condition is : r = 1  and  p=1

put : x= a  and  b = 1  in (*) you have :

(x + 1)^3 = x^3 +3x² +3x +1

the zeros is : x = - 1

bcause : x^3 +3x² +3x +1 = 0     (x + 1)^3 = 0  so : x = - 1

none cuz there are no "following statements" regarding the question

step-by-step explanation:

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)