How do you write a polynomial function of least degree that has real coefficients, the following given zeros: # 1, 1+sqrt(2), 1-sqrt(2)#? Precalculus Polynomial Functions of Higher Degree Zeros 1 Answer Cesareo R. Jul 25, 2016 #x^3 - 3 x^2 + x+1# Explanation: #(x-1-sqrt(2))(x-1+sqrt(2))(x-1) = ((x-1)^2-2)(x-1) # #=x^3 - 3 x^2 + x+1# 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 1439 views around the world You can reuse this answer Creative Commons License