How do you find all the zeros of $x^{3} - 6 x^{2} + 13 x - 10$ $x^3-6x^2+13x-10$ with its multiplicities?

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How do you find all the zeros of $x^{3} - 6 x^{2} + 13 x - 10$ $x^3-6x^2+13x-10$ with its multiplicities?

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Answer 1 · 2016-04-17T11:01:57

$2, 2 \pm i$ $2, 2+-i$

Explanation:

We will use the Rational Root Theorem:

If the rational number r/s is a root of a polynomial whose coefficients are integers, then the integer r is a factor of the constant term, and the integer s is a factor of the leading coefficient.

So, the candidates for roots are:

$\pm 1, \pm 2, \pm 5, \pm 10$ $+-1, +-2,+-5,+-10$

We discover than 2 is a root.

Now we divide the polymorph by (x-2) to discover the other roots:

$\frac{x^{3} - 6 x^{2} + 13 x - 10}{x - 2} = x^{2} + \frac{- 4 x^{2} + 13 x - 10}{x - 2}$ $(x^3-6x^2+13x-10)/(x-2)=x^2+(-4x^2+13x-10)/(x-2)$
$= x^{2} - 4 x + \frac{5 x - 10}{x - 2} = x^{2} - 4 x + 5$ $=x^2-4x+(5x-10)/(x-2)=x^2-4x+5$

This second degree polynom is solved with the quadratic formula:

$x = \frac{4 \pm \sqrt{4^{2} - 4 \cdot 1 \cdot 5}}{2} = \frac{4 \pm \sqrt{16 - 20}}{2} = \frac{4 \pm \sqrt{- 4}}{2} = \frac{4 \pm 2 i}{2} = 2 \pm i$ $x=(4+-sqrt(4^2-4*1*5))/2=(4+-sqrt(16-20))/2=(4+-sqrt(-4))/2=(4+-2i)/2=2+-i$

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How do you find all the zeros of $x^{3} - 6 x^{2} + 13 x - 10$ $x^3-6x^2+13x-10$ with its multiplicities?

1 Answer

Explanation:

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How do you find all the zeros of x3−6x2+13x−10x^3-6x^2+13x-10 with its multiplicities?

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Explanation:

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How do you find all the zeros of $x^{3} - 6 x^{2} + 13 x - 10$ $x^3-6x^2+13x-10$ with its multiplicities?