How do you find all the zeros of $f (x) = 4 x^{3} - 20 x^{2} - 3 x + 15$ $f(x)=4x^3-20x^2-3x+15$ ?

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How do you find all the zeros of $f (x) = 4 x^{3} - 20 x^{2} - 3 x + 15$ $f(x)=4x^3-20x^2-3x+15$ ?

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Answer 1 · 2016-02-27T00:24:58

Factor by grouping and by using the difference of squares identity to find:

$f (x) = (2 x - \sqrt{3}) (2 x + \sqrt{3}) (x - 5)$ $f(x) =(2x-sqrt(3))(2x+sqrt(3))(x-5)$

hence has zeros $x = \pm \frac{\sqrt{3}}{2}$ $x = +-sqrt(3)/2$ and $x = 5$ $x = 5$

Explanation:

Factor by grouping, then use the difference of squares identity:

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

with $a = 2 x$ $a=2x$ and $b = \sqrt{3}$ $b=sqrt(3)$ , as follows:

$f (x) = 4 x^{3} - 20 x^{2} - 3 x + 15$ $f(x) = 4x^3-20x^2-3x+15$

$= (4 x^{3} - 20 x^{2}) - (3 x - 15)$ $=(4x^3-20x^2)-(3x-15)$

$= 4 x^{2} (x - 5) - 3 (x - 5)$ $=4x^2(x-5)-3(x-5)$

$= (4 x^{2} - 3) (x - 5)$ $=(4x^2-3)(x-5)$

$= ({(2 x)}^{2} - {(\sqrt{3})}^{2}) (x - 5)$ $=((2x)^2-(sqrt(3))^2)(x-5)$

$= (2 x - \sqrt{3}) (2 x + \sqrt{3}) (x - 5)$ $=(2x-sqrt(3))(2x+sqrt(3))(x-5)$

So the zeros of $f (x)$ $f(x)$ are $x = \pm \frac{\sqrt{3}}{2}$ $x = +-sqrt(3)/2$ and $x = 5$ $x = 5$

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How do you find all the zeros of $f (x) = 4 x^{3} - 20 x^{2} - 3 x + 15$ $f(x)=4x^3-20x^2-3x+15$ ?

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Explanation:

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How do you find all the zeros of f(x)=4x3−20x2−3x+15f(x)=4x^3-20x^2-3x+15?

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Explanation:

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How do you find all the zeros of $f (x) = 4 x^{3} - 20 x^{2} - 3 x + 15$ $f(x)=4x^3-20x^2-3x+15$ ?