How do you factor $x^{3} - 9 x^{2} + 25 x - 21$ $x^3-9x^2+25x-21$ , given that $x = 3$ $x=3$ is a zero ?

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How do you factor $x^{3} - 9 x^{2} + 25 x - 21$ $x^3-9x^2+25x-21$ , given that $x = 3$ $x=3$ is a zero ?

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Answer 1 · 2016-11-02T19:18:09

$x^{3} - 9 x^{2} + 25 x - 21 = (x - 3) (x - 3 - \sqrt{2}) (x - 3 + \sqrt{2})$ $x^3-9x^2+25x-21 = (x-3)(x-3-sqrt(2))(x-3+sqrt(2))$

Explanation:

$h (x) = x^{3} - 9 x^{2} + 25 x - 21$ $h(x) = x^3-9x^2+25x-21$

The difference of squares identity can be written:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2 = (a-b)(a+b)$

We use this later with $a = (x - 3)$ $a = (x-3)$ and $b = \sqrt{2}$ $b=sqrt(2)$

We are told that $3$ $3$ is a zero, so $(x - 3)$ $(x-3)$ must be a factor:

$x^{3} - 9 x^{2} + 25 x - 21 = (x - 3) (x^{2} - 6 x + 7)$ $x^3-9x^2+25x-21 = (x-3)(x^2-6x+7)$

Then, completing the square we find:

$x^{2} - 6 x + 7 = x^{2} - 6 x + 9 - 2$ $x^2-6x+7 = x^2-6x+9-2$

$x^{2} - 6 x + 7 = {(x - 3)}^{2} - {(\sqrt{2})}^{2}$ $color(white)(x^2-6x+7) = (x-3)^2-(sqrt(2))^2$

$x^{2} - 6 x + 7 = ((x - 3) - \sqrt{2}) ((x - 3) + \sqrt{2})$ $color(white)(x^2-6x+7) = ((x-3)-sqrt(2))((x-3)+sqrt(2))$

$x^{2} - 6 x + 7 = (x - 3 - \sqrt{2}) (x - 3 + \sqrt{2})$ $color(white)(x^2-6x+7) = (x-3-sqrt(2))(x-3+sqrt(2))$

Putting it all together:

$x^{3} - 9 x^{2} + 25 x - 21 = (x - 3) (x - 3 - \sqrt{2}) (x - 3 + \sqrt{2})$ $x^3-9x^2+25x-21 = (x-3)(x-3-sqrt(2))(x-3+sqrt(2))$

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How do you factor $x^{3} - 9 x^{2} + 25 x - 21$ $x^3-9x^2+25x-21$ , given that $x = 3$ $x=3$ is a zero ?

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Explanation:

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How do you factor $x^{3} - 9 x^{2} + 25 x - 21$ $x^3-9x^2+25x-21$ , given that $x = 3$ $x=3$ is a zero ?