Question

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

Answer 1

`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`.