How do you find all the zeros of $f (x) = 2 x^{3} + 3 x^{2} + 8 x - 5$ $f(x)= 2x^3 + 3x^2+ 8x- 5$ ?

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How do you find all the zeros of $f (x) = 2 x^{3} + 3 x^{2} + 8 x - 5$ $f(x)= 2x^3 + 3x^2+ 8x- 5$ ?

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Answer 1 · 2016-08-03T19:31:28

$f (x)$ $f(x)$ has zeros $\frac{1}{2}$ $1/2$ and $- 1 \pm 2 i$ $-1+-2i$

Explanation:

$f (x) = 2 x^{3} + 3 x^{2} + 8 x - 5$ $f(x) = 2x^3+3x^2+8x-5$

By the rational root theorem, any rational zeros of $f (x)$ $f(x)$ are expressible in the form $\frac{p}{q}$ $p/q$ for integers $p, q$ $p, q$ with $p$ $p$ a divisor of the constant term $- 5$ $-5$ and $q$ $q$ a divisor of the coefficient $2$ $2$ of the leading term.

That means that the only possible ratioanl zeros of $f (x)$ $f(x)$ are:

$\pm \frac{1}{2}, \pm 1, \pm \frac{5}{2}, \pm 5$ $+-1/2, +-1, +-5/2, +-5$

We find:

$f (\frac{1}{2}) = \frac{2}{8} + \frac{3}{4} + \frac{8}{2} - 5 = \frac{1}{4} + \frac{3}{4} + 4 - 5 = 0$ $f(1/2) = 2/8+3/4+8/2-5 = 1/4+3/4+4-5 = 0$

So $x = \frac{1}{2}$ $x=1/2$ is a zero of $f (x)$ $f(x)$ and $(2 x - 1)$ $(2x-1)$ a factor:

$2 x^{3} + 3 x^{2} + 8 x - 5$ $2x^3+3x^2+8x-5$

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

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

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

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

Hence the other two zeros are:

$x = - 1 \pm 2 i$ $x = -1+-2i$

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How do you find all the zeros of $f (x) = 2 x^{3} + 3 x^{2} + 8 x - 5$ $f(x)= 2x^3 + 3x^2+ 8x- 5$ ?

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Explanation:

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How do you find all the zeros of f(x)=2x3+3x2+8x−5f(x)= 2x^3 + 3x^2+ 8x- 5?

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Explanation:

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How do you find all the zeros of $f (x) = 2 x^{3} + 3 x^{2} + 8 x - 5$ $f(x)= 2x^3 + 3x^2+ 8x- 5$ ?