Originally Posted by

**skeeter** I'm not familiar with the author's process ... I'd have to go over it if I have time.

I would approach the problem in this manner ...

If it is reducible, the quartic polynomial will have a quadratic square root, i.e. ...

$x^4-6x^3+19x^2-30x+25 = (ax^2+bx+c)^2$

$x^4-6x^3+19x^2-30x+25 = a^2x^2 + 2abx^3 + (2ac+b^2)x^2 + 2bcx + c^2$

matching coefficients, it's easy to see $a=1$ and $c=5$

$2b = -6 \implies b = -3$

so ...

$(x^2-3x+5)^2 = x^4-6x^3+19x^2-30x+25$

now, would $(-x^2 + 3x -5)^2$ also work?