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

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Answer 1 · 2017-01-18T18:56:32

$0$ , $1 + sqrt2$ , and $1 - sqrt2$ .

$0$ 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 $x = 1 - sqrt2$ .

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