How do you find all zeros with multiplicities of $f (x) = 9 x^{3} - 5 x^{2} - x$ $f(x)=9x^3-5x^2-x$ ?

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How do you find all zeros with multiplicities of $f (x) = 9 x^{3} - 5 x^{2} - x$ $f(x)=9x^3-5x^2-x$ ?

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Answer 1 · 2018-08-11T20:05:55

$x = 0$ $x=0" "$ or $x = \frac{5}{18} \pm \frac{\sqrt{61}}{18}$ $" "x = 5/18+-sqrt(61)/18$

all with multiplicity $1$ $1$

Explanation:

$0 = 36 f (x)$ $0 = 36f(x)$

$0 = 36 (9 x^{3} - 5 x^{2} - x)$ $color(white)(0) = 36(9x^3-5x^2-x)$

$0 = x (324 x^{2} - 180 x - 36)$ $color(white)(0) = x(324x^2-180x-36)$

$0 = x ({(18 x)}^{2} - 2 (18 x) (5) + 25 - 61)$ $color(white)(0) = x((18x)^2-2(18x)(5)+25-61)$

$0 = x ({(18 x - 5)}^{2} - {(\sqrt{61})}^{2})$ $color(white)(0) = x((18x-5)^2-(sqrt(61))^2)$

$0 = x (18 x - 5 - \sqrt{61}) (18 x - 5 + \sqrt{61})$ $color(white)(0) = x(18x-5-sqrt(61))(18x-5+sqrt(61))$

Hence:

$x = 0$ $x=0$

or:

$18 x = 5 \pm \sqrt{61}$ $18x=5+-sqrt(61)" "$ so $x = \frac{5}{18} \pm \frac{\sqrt{61}}{18}$ $" "x = 5/18+-sqrt(61)/18$

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How do you find all zeros with multiplicities of $f (x) = 9 x^{3} - 5 x^{2} - x$ $f(x)=9x^3-5x^2-x$ ?

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Explanation:

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How do you find all zeros with multiplicities of f(x)=9x3−5x2−xf(x)=9x^3-5x^2-x?

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How do you find all zeros with multiplicities of $f (x) = 9 x^{3} - 5 x^{2} - x$ $f(x)=9x^3-5x^2-x$ ?