# Proof reading of one college-Algebra-polynomial-thereom.

Welcome to edit my post to revise any mistakes, especially English, thanks.

Theorem 11: If an nozero integer polynomial $f(x)$ can be factored as a product of two rational polynomials and less degree, then it can be factored as a product of two integer polynomials and less degree.

Suppose integer polynomial $f (x)$ has a factorization. $f(x)=g(x) h(x)$, where $f (x)$, $g (x)$ are all rational polynomials,and

$\partial (g (x)) < \partial (f (x))$ $\partial (h (x)) < \partial (f (x))$

Let $f (x) = a f_1 (x)$, $g (x) = r g_1 (x)$, $h (x) = s h_1 (x)$, Here $f_1(x)$, $g_1 (x)$ are all primitive polynomials, where a is an integer, r and s are all rational numbers.

Hence $af_1(x)=r s g_1(x)h_1(x)$.

From Thereom 10(Gaussian Lemma: product of two primitive polynomials is also a primitive polynomial), $g_1(x)h_1(x)$ is primitive polynomial, therefore $r s=\pm a$, This is saying that, $r s$ is an integer. Hence, we have $f(x)=\left(r s g_1(x)\right)h_1(x)$. Here $r\text{ }s g_1(x)$ and $h_1(x)$ are all integer polynomials and with degree less than that of $f(x)$

• – DonAntonio May 28 '13 at 8:58
• The English is ok: there are a few errors (e.g. it should be "are such questions welcome?" and you've spelled Theorem and nonzero wrongly) but it is good enough that we can understand you clearly. – Matthew Towers May 28 '13 at 9:26
• How precisely do you deduce that $\, rs = \pm a\,$? Do your prior theorems handle rational (vs. integral) content? – Key Ideas May 28 '13 at 14:20
• @Key Ideas Theorem 10 is Gauss Lemma, Two primitive polynomials' product is also an primitive polynomial with integer coefficients. – HyperGroups May 28 '13 at 14:32
• @mt_ thanks, I should be more confident in English with your encouragement. Which is right? integer polynomial or integral polynomial – HyperGroups May 31 '13 at 11:09