Question

How do you find the roots of `x^3-5x^2+4x+10=0`?

Answer 1

The roots are: `-1` and `3+-i`

Explanation:

`f(x) = x^3-5x^2+4x+10`

By the reational roots theorem, any rational zeros of `f(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `10` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-2, +-5, +-10`

We find:

`f(-1) = -1-5-4+10 = 0`

So `x=-1` is a zero and `(x+1)` a factor:

`x^3-5x^2+4x+10 = (x+1)(x^2-6x+10)`

We can factor the remaining quadratic by completing the square and using the difference of squares identity:

`a^2-b^2 = (a-b)(a+b)`

with `a=(x-3)` and `b=i` as follows:

`x^2-6x+10 = x^2-6x+9+1`

`color(white)(x^2-6x+10) = (x-3)^2-i^2`

`color(white)(x^2-6x+10) = ((x-3)-i)((x-3)+i)`

`color(white)(x^2-6x+10) = (x-3-i)(x-3+i)`

Hence the other two zeros are: `x = 3+-i`