How do you find all the zeros of $f (x) = x^{3} - 4 x^{2} + 9 x - 36$ $f(x) = x^3-4x^2+9x-36$ ?

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How do you find all the zeros of $f (x) = x^{3} - 4 x^{2} + 9 x - 36$ $f(x) = x^3-4x^2+9x-36$ ?

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Answer 1 · 2018-05-23T22:25:52

The zeros are:

$x = 4$ $x=4" "$ and $x = \pm 3 i$ $" "x=+-3i$

Explanation:

Given:

$f (x) = x^{3} - 4 x^{2} + 9 x - 36$ $f(x) = x^3-4x^2+9x-36$

Note that the ratio between the first and second terms is the same as that between the third and fourth terms. So this cubic will factor by grouping:

$x^{3} - 4 x^{2} + 9 x - 36 = (x^{3} - 4 x^{2}) + (9 x - 36)$ $x^3-4x^2+9x-36 = (x^3-4x^2)+(9x-36)$

$x^{3} - 4 x^{2} + 9 x - 36 = x^{2} (x - 4) + 9 (x - 4)$ $color(white)(x^3-4x^2+9x-36) = x^2(x-4)+9(x-4)$

$x^{3} - 4 x^{2} + 9 x - 36 = (x^{2} + 9) (x - 4)$ $color(white)(x^3-4x^2+9x-36) = (x^2+9)(x-4)$

$x^{3} - 4 x^{2} + 9 x - 36 = (x^{2} + 3^{2}) (x - 4)$ $color(white)(x^3-4x^2+9x-36) = (x^2+3^2)(x-4)$

$x^{3} - 4 x^{2} + 9 x - 36 = (x^{2} - {(3 i)}^{2}) (x - 4)$ $color(white)(x^3-4x^2+9x-36) = (x^2-(3i)^2)(x-4)$

$x^{3} - 4 x^{2} + 9 x - 36 = (x - 3 i) (x + 3 i) (x - 4)$ $color(white)(x^3-4x^2+9x-36) = (x-3i)(x+3i)(x-4)$

So the zeros are:

$x = 4$ $x=4" "$ and $x = \pm 3 i$ $" "x=+-3i$

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How do you find all the zeros of $f (x) = x^{3} - 4 x^{2} + 9 x - 36$ $f(x) = x^3-4x^2+9x-36$ ?

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Explanation:

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How do you find all the zeros of f(x) = x^3-4x^2+9x-36f(x)=x3−4x2+9x−36f(x) = x^3-4x^2+9x-36?

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Explanation:

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How do you find all the zeros of $f (x) = x^{3} - 4 x^{2} + 9 x - 36$ $f(x) = x^3-4x^2+9x-36$ ?