# How do you find the roots of x^3+x+10=0?

Dec 5, 2016

$1 + 2 i$, $1 - 2 i$ and $- 2$

#### Explanation:

We can use the remainder and factor theorems to effectively factor this trinomial.

If $y = a {x}^{n} + b {x}^{n - 1} + \ldots + p$, then the possible factors are $\text{factors of p"/"factors of a}$.

This gives us $\frac{\pm 10 , \pm 5 , \pm 2 , \pm 1}{\pm 1}$. Usually, when I do these sorts of problems, I start at $+ 1$, then work my way up. If you evaluate these factors within the equation and they make the equation true, you have a zero. You will find $x = - 2$ is one of these values.

We now use synthetic division to factor the trinomial.

We want to divide ${x}^{3} + x + 10$ by $x + 2$.

The equation then becomes $\left(x + 2\right) \left({x}^{2} - 2 x + 5\right) = 0$

${x}^{2} - 2 x + 5$ can be solved using the quadratic formula.

$x = \frac{- \left(- 2\right) \pm \sqrt{{\left(- 2\right)}^{2} - 4 \times 1 \times 5}}{2 \times 1}$

$x = \frac{2 \pm \sqrt{- 16}}{2}$

$x = \frac{2 \pm 4 i}{2}$

$x = 1 \pm 2 i$

Putting all of this together, the equation ${x}^{3} + x + 10 = 0$ has roots of $1 + 2 i$, $1 - 2 i$ and $- 2$.

Hopefully this helps!

Dec 5, 2016

$- 2 \mathmr{and} 1 \pm i 2$

#### Explanation:

The number of changes in signs of the coefficients is 0.

So, there are no positive roots.

Checking signs of the cubic, for #x = -1 and -2, we

find that -2 is a zero of the cubic.

So, it is of the form

(x+2)(x^2+ax+b).

Comparing like coefficients,

$a = - 2 \mathmr{and} \mathmr{and} b = 5$. Now, the zeros of the quadratic factor

${x}^{2} - 2 x + 5$ are $1 \pm i 2$