The first thing to do is to write down the coefficients: 1 4 7 16 12.

Clearly, because the signs between them are plus, the sum of the coefficients can't be zero.

For all odd numbered powers of x, insert a minus sign before its coefficient, and put a plus in front of all the others:

1-4+7-16+12=0, so x=-1 is a zero and x+1 is a factor. Use synthetic division to reduce the polynomial:

-1 | 1..4..7 16..12

......1 -1 -3 -4 -12

......1..3..4 12 | 0

The polynomial reduces to x^3+3x^2+4x+12. Now look at 12=2*2*3. So it's possible that x+2 and/or x+3 are factors. This means the zeroes might be -2 and/or -3. First, try -2: -8+12-8+12=8, so -2 is not a zero. Try -3: -27+27-12+12=0, so -3 is a zero and x+3 is a factor. Use synthetic division again to reduce to a quadratic:

-3 | 1..3 4..12

......1 -3 0 -12

......1..0 4 | 0

Now we have: x^2+4 which has no real factors.

Therefore complete factorisation is: (x+1)(x+3)(x^2+4).

The quadratic factorises with imaginary roots 2i and -2i, where i is the imaginary square root of -1. If these are included we have:

(x+1)(x+3)(x+2i)(x-2i)