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

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

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Answer 1 · 2016-08-05T22:10:52

$f (x)$ $f(x)$ has zeros: $1$ $1$ , $- 1$ $-1$ , $- \frac{5}{2}$ $-5/2$

Explanation:

This cubic factors by grouping:

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

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

$= x^{2} (2 x + 5) - 1 (2 x + 5)$ $=x^2(2x+5)-1(2x+5)$

$= (x^{2} - 1) (2 x + 5)$ $=(x^2-1)(2x+5)$

$= (x^{2} - 1^{2}) (2 x + 5)$ $=(x^2-1^2)(2x+5)$

$= (x - 1) (x + 1) (2 x + 5)$ $=(x-1)(x+1)(2x+5)$

Hence zeros: $1$ $1$ , $- 1$ $-1$ , $- \frac{5}{2}$ $-5/2$

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

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How do you find all the zeros of f(x)=2x3+5x2−2x−5f(x)=2x^3+5x^2-2x-5?

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Explanation:

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