How do I find a function in standard form that has the zeros 4,sqrt2,-sqrt2 ?

Answer 1

f(x) = (x-4)(x-√2)(x+√2)

(x-√2) and (x+√2) are conjugates, so their product is x²-2. From (a+b)(a-b) = a²-b².

f(x) = (x-4)(x²-2) = x³-2x-4x²+8

= x³-4x²-2x+8

Answer 2

f(x) = (x-4)(x^2-2)