Question

How do you factor `x^3 - 4x^2 - 11x + 2 = 0`?

Answer 1

`x^3-4x^2-11x+2 = (x+2)(x-3-2sqrt(2))(x-3+2sqrt(2))`

Explanation:

`f(x) = x^3-4x^2-11x+2`

By the rational roots 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 `2` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-2`

Trying each in turn, we find:

`f(-2) = -8-4(4)+11(2)+2 = -8-16+22+2 = 0`

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

`x^3-4x^2-11x+2 = (x+2)(x^2-6x+1)`

Factor the remaining quadratic by completing the square and using the difference of squares identity:

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

as follows:

`x^2-6x+1`

`=x^2-6x+9-8`

`=(x-3)^2-(2sqrt(2))^2`

`=((x-3)-2sqrt(2))((x-3)+2sqrt(2))`

`=(x-3-2sqrt(2))(x-3+2sqrt(2))`

Putting it all together:

`x^3-4x^2-11x+2 = (x+2)(x-3-2sqrt(2))(x-3+2sqrt(2))`