How do you factor `x^4 + 4x^3 - 22x^2 + 4x + 16`?

`f(x) = x^4+4x^3-22x^2+4x+16`

We can try the rational roots theorem, which tells us that any rational zeros of `f(x)` are expressibl in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `16` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-2, +-4, +-8, +-16`

None of these works, so `f(x)` has no rational zeros and therefore no linear factors with integer coefficients.

How about quadratic factors?

Is there a pair of quadratic factors with integer coefficients?

We can simplify the problem by making a linear substitution.

`x^4+4x^3-22x^2+4x+16`

`=(x+1)^4-28(x+1)^2+56(x+1)-13`

`=t^4-28t^2+56t-13`

where `t=x+1`

If this simplified quartic has quadratic factors with integer coefficients then they must take one of the following forms:

• `(t^2+at+13)(t^2-at-1)`

• `(t^2+at-13)(t^2-at+1)`

With a little comparing of coefficients we find:

`t^4-28t^2+56t-13 = (t^2+4t-13)(t^2-4t+1)`

Hence:

`x^4+4x^3-22x^2+4x+16`

`=((x+1)^2+4(x+1)-13)((x+1)^2-4(x+1)+1)`

`=(x^2+6x-8)(x^2-2x-2)`

`=((x+3)^2-17)((x-1)^2-3)`

`=(x+3-sqrt(17))(x+3+sqrt(17))(x-1-sqrt(3))(x-1+sqrt(3))`