How do you write a polynomial function of least degree and leading coefficient 1 when the zeros are 2i, -2i? Precalculus Polynomial Functions of Higher Degree Zeros 1 Answer Alan N. Sep 14, 2016 f(x) = x^2+4f(x)=x2+4 Explanation: Let f(x)f(x) be the polynomial of degree 2 and leading coefficient 1 with zeros +-2i±2i Hence either (x+2i)(x+2i) or (x-2i) = 0 (x−2i)=0 :. f(x) = 0 = (x+2i)(x-2i) f(x) = x^2 +2ix -2ix -4i^2 f(x) = x^2 +4 Answer link Related questions What is a zero of a function? How do I find the real zeros of a function? How do I find the real zeros of a function on a calculator? What do the zeros of a function represent? What are the zeros of f(x) = 5x^7 − x + 216? What are the zeros of f(x)= −4x^5 + 3? How many times does f(x)= 6x^11 - 3x^5 + 2 intersect the x-axis? What are the real zeros of f(x) = 3x^6 + 1? How do you find the roots for 4x^4-26x^3+50x^2-52x+84=0? What are the intercepts for the graphs of the equation y=(x^2-49)/(7x^4)? See all questions in Zeros Impact of this question 1349 views around the world You can reuse this answer Creative Commons License