0
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ If you allow real coefficients, you can factorize further : $3(x-\frac{1}{3})(x-\sqrt{2})(x+\sqrt{2})(x^2+1)$. $\endgroup$ Nov 1, 2020 at 21:16
  • $\begingroup$ 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 $\bf C$). $\endgroup$
    – Bernard
    Nov 1, 2020 at 21:17

1 Answer 1

1
$\begingroup$

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 $$3x^5 - x^4 - 3x^3 + x^2 - 6x + 2 = 3(x - 1/3)(x - \sqrt{2})(x + \sqrt{2})(x + i)(x - i).$$

$\endgroup$
2
  • 2
    $\begingroup$ For your second case, do you mean reals rather than rationals? $\sqrt2$ is not rational. $\endgroup$
    – badjohn
    Nov 1, 2020 at 21:29
  • $\begingroup$ @badjohn No, it should've been reals. $\endgroup$ Nov 1, 2020 at 21:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.