How do you find all the real and complex roots of $x^{3} + 9 x^{2} + 19 x - 29 = 0$ $x^3 + 9x^2 + 19x - 29 = 0$ ?

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How do you find all the real and complex roots of $x^{3} + 9 x^{2} + 19 x - 29 = 0$ $x^3 + 9x^2 + 19x - 29 = 0$ ?

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Answer 1 · 2016-01-22T16:40:30

$x = 1, - 5 \pm 2 i$ $x=1,-5+-2i$

Explanation:

The potential real roots of a polynomial are the factors of $\frac{p}{q}$ $p/q$ , where $p$ $p$ is the constant and $q$ $q$ is the coefficient of the term with the highest degree.

Since $p = 29$ $p=29$ and $q = 1$ $q=1$ , the possible real root(s) must be factors of $\frac{29}{1} = 29$ $29/1=29$ . This makes the problem significantly easier since the $29$ $29$ is prime, so the only possibilities for the roots are $\pm 1, \pm 29$ $+-1,+-29$ .

Use synthetic division or polynomial long division to find that $(x - 1)$ $(x-1)$ is a factor, which means that $1$ $color(red)1$ is a root:

$\frac{x^{3} + 9 x^{2} + 19 x - 29}{x - 1} = x^{2} + 10 x + 29$ $(x^3+9x^2+19x-29)/(x-1)=x^2+10x+29$

The other two roots can be found through applying the quadratic formula on $x^{2} + 10 x + 29$ $x^2+10x+29$ :

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} = \frac{- 10 \pm \sqrt{- 16}}{2} = - 5 \pm 2 i$ $x=(-b+-sqrt(b^2-4ac))/(2a)=(-10+-sqrt(-16))/2=color(red)(-5+-2i$

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How do you find all the real and complex roots of $x^{3} + 9 x^{2} + 19 x - 29 = 0$ $x^3 + 9x^2 + 19x - 29 = 0$ ?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you find all the real and complex roots of x^3 + 9x^2 + 19x - 29 = 0x3+9x2+19x−29=0x^3 + 9x^2 + 19x - 29 = 0?

1 Answer

Explanation:

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Impact of this question

How do you find all the real and complex roots of $x^{3} + 9 x^{2} + 19 x - 29 = 0$ $x^3 + 9x^2 + 19x - 29 = 0$ ?