How do you use the rational roots theorem to find all possible zeros of $x^{4} - 9 x^{2} + 20$ $x^4-9x^2+20$ ?

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How do you use the rational roots theorem to find all possible zeros of $x^{4} - 9 x^{2} + 20$ $x^4-9x^2+20$ ?

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Answer 1 · 2016-07-28T21:25:53

Zeros: $x = \pm 2$ $x = +-2$ and $x = \pm \sqrt{5}$ $x = +-sqrt(5)$

Explanation:

Given $f (x) = x^{4} - 9 x^{2} + 20$ $f(x) = x^4-9x^2+20$ , the rational roots theorem tells you that any rational zeros must be expressible in the form $\frac{p}{q}$ $p/q$ for integers $p$ $p$ , $q$ $q$ with $p$ $p$ a divisor of the constant term $20$ $20$ and $q$ $q$ a divisor of the coefficient $1$ $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20$ $+-1, +-2, +-4, +-5, +-10, +-20$

Evaluating $f (x)$ $f(x)$ for each in turn, this allows you to find the two rational zeros:

$f (2) = f (- 2) = 16 - 36 + 20 = 0$ $f(2) = f(-2) = 16-36+20 = 0$

You can then divide $f (x)$ $f(x)$ by $(x - 2) (x + 2) = x^{2} - 4$ $(x-2)(x+2) = x^2-4$ to get $x^{2} - 5$ $x^2-5$ , which has zeros $x = \pm \sqrt{5}$ $x = +-sqrt(5)$ .

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Alternatively:

Note that $x^{4} - 9 x^{2} + 20$ $x^4-9x^2+20$ is a quadratic in $x^{2}$ $x^2$ . That is:

$x^{4} - 9 x^{2} + 20 = {(x^{2})}^{2} - 9 (x^{2}) + 20$ $x^4-9x^2+20 = (x^2)^2-9(x^2)+20$

Note also that $4 + 5 = 9$ $4+5 = 9$ and $4 \times 5 = 20$ $4xx5 = 20$

So we find:

$x^{4} - 9 x^{2} + 20$ $x^4-9x^2+20$

$= (x^{2} - 4) (x^{2} - 5)$ $=(x^2-4)(x^2-5)$

$= (x^{2} - 2^{2}) (x^{2} - {(\sqrt{5})}^{2})$ $=(x^2-2^2)(x^2-(sqrt(5))^2)$

$= (x - 2) (x + 2) (x - \sqrt{5}) (x + \sqrt{5})$ $=(x-2)(x+2)(x-sqrt(5))(x+sqrt(5))$

Hence zeros:

$x = \pm 2$ $x = +-2$ and $x = \pm \sqrt{5}$ $x = +-sqrt(5)$

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How do you use the rational roots theorem to find all possible zeros of $x^{4} - 9 x^{2} + 20$ $x^4-9x^2+20$ ?

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