How do you find all the zeros of $f (x) = 2 x^{3} - x^{2} - 3 x - 1$ $f(x)=2x^3-x^2-3x-1$ ?

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How do you find all the zeros of $f (x) = 2 x^{3} - x^{2} - 3 x - 1$ $f(x)=2x^3-x^2-3x-1$ ?

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Answer 1 · 2016-08-02T23:37:28

Zeros: $- \frac{1}{2}$ $-1/2$ and $\frac{1}{2} \pm \frac{\sqrt{5}}{2}$ $1/2+-sqrt(5)/2$

Explanation:

$f (x) = 2 x^{3} - x^{2} - 3 x - 1$ $f(x) = 2x^3-x^2-3x-1$

By the rational root theorem any rational zeros of $f (x)$ $f(x)$ must be expressible in the form $\frac{p}{q}$ $p/q$ for integers $p, q$ $p, q$ with $p$ $p$ a divisor of the constant term $- 1$ $-1$ and $q$ $q$ a divisor of the coefficient $2$ $2$ of the leading term.

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

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

We find:

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

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

$2 x^{3} - x^{2} - 3 x - 1$ $2x^3-x^2-3x-1$

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

$= (2 x + 1) ({(x - \frac{1}{2})}^{2} - \frac{5}{4})$ $= (2x+1)((x-1/2)^2-5/4)$

$= (2 x + 1) ⎛ ⎝ {(x - \frac{1}{2})}^{2} - {(\frac{\sqrt{5}}{2})}^{2} ⎞ ⎠$ $= (2x+1)((x-1/2)^2-(sqrt(5)/2)^2)$

$= (2 x + 1) (x - \frac{1}{2} - \frac{\sqrt{5}}{2}) (x - \frac{1}{2} + \frac{\sqrt{5}}{2})$ $= (2x+1)(x-1/2-sqrt(5)/2)(x-1/2+sqrt(5)/2)$

Hence the other two zeros are:

$x = \frac{1}{2} \pm \frac{\sqrt{5}}{2}$ $x = 1/2+-sqrt(5)/2$

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How do you find all the zeros of $f (x) = 2 x^{3} - x^{2} - 3 x - 1$ $f(x)=2x^3-x^2-3x-1$ ?

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How do you find all the zeros of f(x)=2x3−x2−3x−1f(x)=2x^3-x^2-3x-1?

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How do you find all the zeros of $f (x) = 2 x^{3} - x^{2} - 3 x - 1$ $f(x)=2x^3-x^2-3x-1$ ?