How do you find all zeros of $f (x) = x^{5} + x^{3} - 6 x$ $f(x)=x^5+x^3-6x$ ?

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Answer 1 · 2017-01-12T07:48:02

The zeros of $f (x)$ $f(x)$ are: $0, \pm \sqrt{2}, \pm \sqrt{3} i$ $0, +-sqrt2, +-sqrt3 i$

$f (x) = x^{5} + x^{3} - 6 x$ $f(x) = x^5+x^3-6x$

The zeros of $f (x)$ $f(x)$ are the values of $x$ $x$ where $f (x) = 0$ $f(x) = 0$

That is where: $x^{5} + x^{3} - 6 x = 0$ $x^5+x^3-6x =0$

$x (x^{4} + x^{2} - 6) = 0$ $x(x^4+x^2-6) = 0$

Hence $x = 0$ $x=0$ or $x^{4} + x^{2} - 6 = 0$ $x^4+x^2-6 =0$

Let $z = x^{2}$ $z=x^2$

$:. z^2 +z -6 =0$

$(z+3)(z-2)=0$

$-> z= 2 or -3$

$:. x=+-sqrt2 or +-sqrt3 i$

The zeros of $f(x)$ are: $0, +-sqrt2, +-sqrt3 i$

How do you find all zeros of f(x)=x^5+x^3-6xf(x)=x5+x3−6xf(x)=x^5+x^3-6x?