How do you find the roots of $x^{3} + x^{2} - 7 x + 2 = 0$ $x^3+x^2-7x+2=0$ ?

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How do you find the roots of $x^{3} + x^{2} - 7 x + 2 = 0$ $x^3+x^2-7x+2=0$ ?

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Answer 1 · 2016-11-11T07:27:50

The roots are $2$ $2$ and $- \frac{3}{2} \pm \frac{\sqrt{13}}{2}$ $-3/2 +- sqrt(13)/2$

Explanation:

Given:

$f (x) = x^{3} + x^{2} - 7 x + 2$ $f(x) = x^3+x^2-7x+2$

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 $2$ $2$ and $q$ $q$ a divisor of the coefficient $1$ $1$ of the leading term.

That means that the only possible rational zeros are:

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

Note that:

$f (2) = 8 + 4 - 14 + 2 = 0$ $f(2) = 8+4-14+2 = 0$

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

$x^{3} + x^{2} - 7 x + 2 = (x - 2) (x^{2} + 3 x - 1)$ $x^3+x^2-7x+2 = (x-2)(x^2+3x-1)$

We can find the zeros of the remaining quadratic factor using the quadratic formula with $a = 1$ $a=1$ , $b = 3$ $b=3$ and $c = - 1$ $c=-1$ :

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

$x = \frac{- 3 \pm \sqrt{9 + 4}}{2}$ $color(white)(x) = (-3+-sqrt(9+4))/2$

$x = - \frac{3}{2} \pm \frac{\sqrt{13}}{2}$ $color(white)(x) = -3/2+-sqrt(13)/2$

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How do you find the roots of $x^{3} + x^{2} - 7 x + 2 = 0$ $x^3+x^2-7x+2=0$ ?

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Explanation:

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How do you find the roots of $x^{3} + x^{2} - 7 x + 2 = 0$ $x^3+x^2-7x+2=0$ ?