How do you find all zeros of #g(x)=5x(x^2-2x-1)#?

1 Answer
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#.