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

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

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Answer 1 · 2017-01-18T18:56:32

$0$ $0$ , $1 + \sqrt{2}$ $1 + sqrt2$ , and $1 - \sqrt{2}$ $1 - sqrt2$ .

Explanation:

$0$ $0$ is clearly a root, since $g (x)$ $g(x)$ is a product that contains $5 x$ $5x$ .

$x^{2} - 2 x - 1$ $x^2 -2x -1$ is a polynomial with discriminant:

$D = b^{2} - 4 a c$ $D = b^2 -4ac$ , where $a, b, c$ $a,b,c$ the coefficients of the polynomial ( $a$ $a$ for $x^{2}$ $x^2$ , $b$ $b$ for $x$ $x$ , $c$ $c$ for the constant). So,

$D = 4 + 4 = 8 > 0$ $D = 4 + 4 = 8 > 0$ and it follows that the aforementioned polynomial has two unique roots given by:

$x = \frac{- b \pm \sqrt{D}}{2 a} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2 \sqrt{2}}{2} = 1 \pm \sqrt{2} .$ $x = (-b +- sqrtD)/(2a) = (2+-sqrt8)/2 = (2 +-2sqrt2)/2 = 1 +- sqrt2.$

So, $g$ $g$ has three roots: $x = 0$ $x = 0$ , $x = 1 + \sqrt{2}$ $x = 1 + sqrt2$ , and $x = 1 - \sqrt{2}$ $x = 1 - sqrt2$ .

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

1 Answer

Explanation:

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How do you find all zeros of g(x)=5x(x2−2x−1)g(x)=5x(x^2-2x-1)?

1 Answer

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