Find a polynomial with degree 3; zeros - 3,3,5

Step-by-step explanation:

f (x) = ax^3+bx^2+cx+d

f (-3) = -3a^3-3b^2-3c+d=0

f (3) = 3a^3+3b^2+3c+d=0

f (5) = 5a^3+5b^2+5c+d=0

f (0) = 0a^3+0b^2+0c+d=2 < = any number you want your extremum at

=> d=2

f (-3) = -3a^3-3b^2-3c+=-2

f (3) = 3a^3+3b^2+3c+=-2

f (5) = 5a^3+5b^2+5c+=-2

solve for all variables (Gauß)

a = 0.0266

b = - 0.08

c = - 0.66

As Solution

0.0266x^3-0.08x^2-0.66x+2 approximately

(x^3) - (5x^2) - 9x+45

Explanation:

(x-3) (x+3) (x-5)

(x^2-9) (x-5)

(x^3) - (5x^2) - 9x+45