# Factoring with integer coefficients vs factoring with real coefficients

What is the difference between factoring with integer coefficients and factoring with real coefficients? Although the answer to this question probably sounds pretty obvious, I'm not sure how my answer changes in the question below.

If I was to factor $$3x^5-x^4-3x^3+x^2-6x+2$$ with integer coefficients, the answer is $$(3x-1)(x^2-2)(x^2+1)$$ How does my answer change if I am factoring with real (not necessarily integer) coefficients?

• If you allow real coefficients, you can factorize further : $3(x-\frac{1}{3})(x-\sqrt{2})(x+\sqrt{2})(x^2+1)$. Nov 1, 2020 at 21:16
• Only $x^2-2$ would change to $(x-\sqrt 2)(x+\sqrt 2)$ (and $x^2+1$ to $(x-i)(x+i)$ if you factor with coefficients in $\bf C$). Nov 1, 2020 at 21:17

Factoring over the integers, we see that $$3x^5 - x^4 - 3x^3 + x^2 - 6x + 2 = (3x - 1)(x^2 - 2)(x^2 + 1).$$ Factoring over the reals, we see that $$3x^5 - x^4 - 3x^3 + x^2 - 6x + 2 = 3(x - 1/3)(x - \sqrt{2})(x + \sqrt{2})(x^2 + 1).$$ Factoring over the complex numbers, we have $$3x^5 - x^4 - 3x^3 + x^2 - 6x + 2 = 3(x - 1/3)(x - \sqrt{2})(x + \sqrt{2})(x + i)(x - i).$$
• For your second case, do you mean reals rather than rationals? $\sqrt2$ is not rational. Nov 1, 2020 at 21:29