What are all the possible rational zeros for $f (x) = x^{3} + 5 x^{2} - 11 x - 6$ $f(x)=x^3+5x^2-11x-6$ and how do you find all zeros?

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What are all the possible rational zeros for $f (x) = x^{3} + 5 x^{2} - 11 x - 6$ $f(x)=x^3+5x^2-11x-6$ and how do you find all zeros?

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Answer 1 · 2016-10-26T20:56:59

$f (x)$ $f(x)$ has zeros $2$ $2$ and $- \frac{7}{2} \pm \frac{\sqrt{37}}{2}$ $-7/2+-sqrt(37)/2$

Explanation:

$f (x) = x^{3} + 5 x^{2} - 11 x - 6$ $f(x) = x^3+5x^2-11x-6$

By the rational roots 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 $6$ $6$ and $q$ $q$ a divisor of the coefficient $1$ $1$ of the leading term.

Thats means that the only possible rational zeros are:

$\pm 1, \pm 2, \pm 3, \pm 6$ $+-1, +-2, +-3, +-6$

In addition, note that the signs of the coefficients of $f (x)$ $f(x)$ are in the pattern: $+ + - -$ $+ + - -$ . With one change of sign, Descartes' Rule of Signs tells us that $f (x)$ $f(x)$ has exactly one positive Real zero. So let's try the positive rational possibilities first:

$f (1) = 1 + 5 - 11 + 6 = 1$ $f(1) = 1+5-11+6 = 1$

$f (2) = 8 + 5 (4) - 11 (2) - 6 = 8 + 20 - 22 - 6 = 0$ $f(2) = 8+5(4)-11(2)-6 = 8+20-22-6 = 0$

So $x = 2$ $x=2$ is a zero and $(x - 2)$ $(x-2)$ a factor of $f (x)$ $f(x)$ :

$x^{3} + 5 x^{2} - 11 x - 6 = (x - 2) (x^{2} + 7 x + 3)$ $x^3+5x^2-11x-6 = (x-2)(x^2+7x+3)$

The remaining two zeros are roots of the quadratic: $x^{2} + 7 x + 3 = 0$ $x^2+7x+3 = 0$ , which we can solve using the quadratic formula:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b+-sqrt(b^2-4ac))/(2a)$

$x = \frac{- 7 \pm \sqrt{7^{2} - 4 (1) (3)}}{2 \cdot 1}$ $color(white)(x) = (-7+-sqrt(7^2-4(1)(3)))/(2*1)$

$x = \frac{- 7 \pm \sqrt{49 - 12}}{2}$ $color(white)(x) = (-7+-sqrt(49-12))/2$

$x = - \frac{7}{2} \pm \frac{\sqrt{37}}{2}$ $color(white)(x) = -7/2+-sqrt(37)/2$

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What are all the possible rational zeros for $f (x) = x^{3} + 5 x^{2} - 11 x - 6$ $f(x)=x^3+5x^2-11x-6$ and how do you find all zeros?

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Explanation:

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What are all the possible rational zeros for f(x)=x^3+5x^2-11x-6f(x)=x3+5x2−11x−6f(x)=x^3+5x^2-11x-6 and how do you find all zeros?

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What are all the possible rational zeros for $f (x) = x^{3} + 5 x^{2} - 11 x - 6$ $f(x)=x^3+5x^2-11x-6$ and how do you find all zeros?