Question

How do you find all the zeros of `f(x) = x^3 - 4x^(2) + 14x - 20`?

Answer 1

`x = 2` or `x = 1+-3i`

Explanation:

`f(x) = x^3-4x^2+14x-20`

By the rational root theorem, any rational zeros of `f(x)` must be expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-20` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1`, `+-2`, `+-4`, `+-5`, `+-10`, `+-20`

In addition note that if you invert the signs of the coefficients of the terms of odd degree, then the pattern of signs you get is:

`- - - -`

which has no changes. So there are no negative zeros.

So the only possible rational zeros are:

`1, 2, 4, 5, 10, 20`

We find:

`f(2) = 8-16+28-20 = 0`

So `x=2` is a zero and `(x-2)` a factor:

`x^3-4x^2+14x-20 = (x-2)(x^2-2x+10)`

The remaining quadratic factor has negative discriminant, so no Real zeros, but we can still factor it by completing the square and using the difference of squares identity:

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

with `a=(x-1)` and `b=3i` as follows:

`x^2-2x+10`

`=x^2-2x+1+3^2`

`=(x-1)^2-(3i)^2`

`=((x-1)-3i)((x-1)+3i)`

`=(x-1-3i)(x-1+3i)`

Hence Complex zeros:

`x = 1+-3i`