How do you find all zeros of #g(x)=5x(x^2-2x-1)#? Precalculus Polynomial Functions of Higher Degree Zeros 1 Answer Azimet Jan 18, 2017 Answer: #0#, #1 + sqrt2#, and #1 - sqrt2#. Explanation: #0# is clearly a root, since #g(x)# is a product that contains #5x#. #x^2 -2x -1# is a polynomial with discriminant: #D = b^2 -4ac#, where #a,b,c# the coefficients of the polynomial (#a# for #x^2#, #b# for #x#, #c# for the constant). So, #D = 4 + 4 = 8 > 0# and it follows that the aforementioned polynomial has two unique roots given by: #x = (-b +- sqrtD)/(2a) = (2+-sqrt8)/2 = (2 +-2sqrt2)/2 = 1 +- sqrt2.# So, #g# has three roots: #x = 0#, #x = 1 + sqrt2#, and #x = 1 - sqrt2#. 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 525 views around the world You can reuse this answer Creative Commons License