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

it is b)

hope this : )

step-by-step explanation:

yes because there is exactly one y value for every x value

for function there can't be more than one value for every x value