Question

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

Answer 1

`1 + 2i`, `1 - 2i` and `-2`

Explanation:

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

If `y = ax^n + bx^(n - 1) + ... + p`, then the possible factors are `"factors of p"/"factors of a"`.

This gives us `(+-10, +-5, +- 2, +-1)/(+- 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 `(x + 2)(x^2 - 2x + 5) = 0`

`x^2 - 2x + 5` can be solved using the quadratic formula.

`x= (-(-2) +- sqrt((-2)^2- 4 xx 1 xx 5))/(2 xx 1)`

`x = (2 +- sqrt(-16))/2`

`x = (2+- 4i)/2`

`x = 1 +- 2i`

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

Hopefully this helps!

Answer 2

`-2 and 1+-i2`

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 and and b = 5`. Now, the zeros of the quadratic factor

`x^2-2x+5` are `1+-i2`