Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am reading a definition in my Pre-Calculus book but I am a little but confused, the definition states:
Suppose $p$ is a nonzero polynomial with at least one (real) zero. Then

*There exist real numbers $r_1$,$r_2$,...,$r_m$ and a polynomial G such that G has no (real) zeroes and $p(x)=(x-r_1)(x-r_2)...(x-r_m)G(x)$ for every real number $x$;

*each of the numbers $r_1$,$r_2$,...,$r_m$ is a zero of $p$;

*$p$ has no zeros other than $r_1$,$r_2$,...,$r_m$.

I understand why $p(x)=(x-r_1)(x-r_2)...(x-r_m)$ makes sense but I am having trouble understanding why there is a polynomial $G(x)$ that has no (real) zeroes at the end. Could someone please explain this to me? I am really confused.

share|improve this question
There are irreducible quadratics in the reals. For example, you can be left over with a factor of $x^2 + 1$. –  EuYu Oct 2 '12 at 3:32

2 Answers 2

up vote 5 down vote accepted

For example: $$x^4 - 1 = (x-1)(x+1)(x^2+1)$$ The real zeroes of $x^4-1$ are 1 and -1. What's left is $x^2+1$, which has no real zeroes.

If $G$ still had another real zero at a point $x=a$, then you could pull out another factor $(x-a)$. So after you factor out all the real zeroes of $p$, what's left cannot have any real zeroes.

share|improve this answer

In general, given a polynomial with real coefficients, you will have a set of real roots and a set of complex conjugate pair roots. For example, you have the polynomial $$x^2 + 1 = (x-i)(x+i)$$ which factors with the conjugate root pair $\pm i$ but the polynomial is irreducible over the real numbers. To see this more clearly, suppose you factor the polynomial over the complex numbers. You will end up with something akin to $$p(x) = [(x-r_1)\cdots(x-r_m)]\cdot[(x-z_1)(x-\overline{z_1})\cdots(x-{z_k})(x-\overline{z_k})]$$ The $r_i$ are your real roots and the $z_i, \overline{z_i}$ are your complex conjugate root pairs (that complex roots always come in such pairs for real coefficient polynomials is a consequence of the conjugate root theorem). Each of the pairs $$(x-z_i)(x-\overline{z_i}) = x^2 - 2\mathrm{Re}(z_i) + |z_i|^2$$ will then multiply to produce an irreducible real quadratic. The collection of these irreducible quadratics will then form your remaining polynomial $G$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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