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

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

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Answer 1 · 2016-01-22T16:40:30

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

Explanation:

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

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

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

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

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

$x=(-b+-sqrt(b^2-4ac))/(2a)=(-10+-sqrt(-16))/2=color(red)(-5+-2i$

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

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 = 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 $x^3 + 9x^2 + 19x - 29 = 0$ ?