# Integer polynomial and division by binomial [duplicate]

If I have an integer coefficient polynomial $$f(x)$$ and know that it has 2 integer roots, $$x_1$$ and $$x_2$$, is it true that $$f(x) = (x-x_1)(x-x_2) h(x)$$ where $$h(x)$$ is another integer coefficient polynomial? What is this result called if it is true? How can I prove that?

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• One approach to proving this (it is true) is by synthetic division as taught in high school/gymnasium. – hardmath Oct 20 at 22:13
• $f(x+x_1)$ has constant coefficient equal to $0$. Therefore, you can factor out one $x$ to get $f(x+x_1)=xg(x)$. Here $f(x+x_1)$ and $g(x)$ have the same collection of coefficients, just shifted one degree lower. Therefore $f(x)=(x-x_1)g(x-x_1)$ Next you can do the same with $g(x-x_1)$ and look at $g(x-x_1+x_2)$. – conditionalMethod Oct 20 at 22:16
• Could you explain in a bit more detail, @hardmath? – Gordon Oct 20 at 22:16
• Sure, but would you indicate what level of math studies you are involved in? – hardmath Oct 20 at 22:18
• I have a PhD in physics, but it's been a long time since I was involved in high school math, as it seems. I am just trying to remind myself some things while helping my brother, so if you can provide some references using high school math, that would be great. – Gordon Oct 20 at 22:19

Let's assume that $$x_1,x_2$$ are distinct integer roots of integer polynomial $$f(x)$$. In that case we can first divide $$f(x)$$ by $$(x-x_1)$$, then divide the resulting quotient by $$(x-x_2)$$ to get the desired result. So the hard part of the proof is showing that the first quotient is again an integer polynomial:

$$f(x) = (x-x_1)g(x)$$

Since $$x_1\neq x_2$$, in order for $$f(x_2) = 0$$, it must be that $$g(x_2) = 0$$. This allows us to do the "division" by $$(x-x_2)$$ just as we will show the first step, and if $$g(x) = (x-x_2) h(x)$$, then:

$$f(x) = (x-x_1)(x-x_2)h(x)$$

With that in mind we will go deeper into the details of why $$g(x)$$ as the first quotient is an integer polynomial.

The topic of synthetic division is generally covered in Algebra II for high school students where the divisor is binomial $$x-r$$ and the dividend $$f(x)$$ is an integer polynomial. It can be generalized to some extent, but this is already the case our immediate purpose requires.

The computation is equivalent to carrying out a long division of polynomials, but the special form of the divisor $$x-r$$ allows for less writing and less arithmetic. Further the equivalence of the synthetic division and the long division allow us to realize that the final remainder $$d$$ of the computation:

$$f(x) = (x-r)g(x) + D$$

is exactly $$D = f(r)$$, the evaluation of polynomial $$f(x)$$ at $$x=r$$.

The main idea is to "attack" the leading coefficient of $$f(x)$$ by subtracting a multiple of $$(x-r)$$ that eliminates that term, thus reducing the degree of the (partial) dividend. Suppose that $$f(x)$$ is an integer polynomial of degree $$n$$:

$$f(x) = A_0 x^n + A_1 x^{n-1} + \ldots + A_{n-1}x + A_n$$

Then we could subtract $$(x-r)\cdot A_0 x^{n-1}$$ from $$f(x)$$ and get a "new" dividend of degree (at least) one less:

$$f(x) - (x-r)\cdot A_0 x^{n-1} = (A_1 + rA_0) x^{n-1} + A_2 x^{n-2} + \ldots + A_{n-1}x + A_n$$

We continue to do this subtraction of multiples of $$(x-r)$$ times the indicated coefficient of smaller and smaller powers of $$x$$ until at last we have removed all but a final constant remainder $$D$$. Piecing everything together we have $$f(x) = (x-r)g(x) + D$$ as promised, where the degree of the combined quotient $$g(x)$$ is one less than the degree of $$f(x)$$.

Notice that although division is being carried, the adjustments of the coefficients at each stage are actually additions of integers. So by induction not only is the final remainder $$D$$ an integer, so are the coefficients of the pieced together quotient $$g(x)$$.

This in brief establishes that if $$r$$ is a root of $$f(x)$$, then the final remainder $$D$$ is zero (because $$D=f(r)$$ as explained earlier), and we have our factorization of integer polynomials:

$$f(x) = (x-r)g(x)$$

Let me know if this much detail is not enough, or if more explanation is needed to clarify what has already been sketched.

Since $$f$$ is an integer coefficient polynomial that has $$2$$ integer roots $$x_1$$ and $$x_2$$, it is divisible by $$(x-x_1)$$ and $$(x-x_2)$$, meaning that it can be rewritten as $$(x-x_1)(x-x_2)h(x)$$, where $$h(x)$$ is another integer coefficient polynomial. This is because $$(x-x_1)(x-x_2)$$ is a monic polynomial, so it doesn't change the leading coefficient of $$f(x)$$. This result is a factorization of $$f(x)$$.

• Okay, but why does $h(x)$ have integer coefficients? – Gordon Oct 20 at 22:21
• @Gordon, because $(x-x_1)(x-x_2)$ is monic – lhf Oct 20 at 22:22

Consider the polynomial $$g_1(x)=f(x+x_1)$$. Its coefficients are integer because to compute them you only need to multiply and add coefficients of $$f$$ and $$x_1$$.

Since $$f(x_1)=0$$, it follows that $$g_1(0)=0$$. This is, the coefficient of $$g$$ of degree zero is zero. Therefore, $$g_1(x)=xg_2(x)$$. The coefficients of $$g_2$$ are the same as the coefficients of $$g_1$$, just shifted to one degree less. Hence, they are also integers.

Replacing $$x$$ by $$x-x_1$$ in the equation $$g_1(x)=xg_2(x)$$ we get $$f(x)=g_1(x-x_1)=(x-x_1)g_2(x-x_1)$$

The coefficients of $$g_3(x)=g_2(x-x_1)$$ are integers because, as before, they are obtained from those of $$g_2$$ by doing only additions and multiplications of them with $$x_1$$.

Repeating the same argument with $$x_2$$ and $$g_3(x)$$ you get the result.

Aside: Above there were two moments in the argument in which one needs to write polynomials like $$f(x-x_1)$$ back into powers of $$x$$, too get its coefficients. This consists in opening all parentheses. However, this can be done, and faster, by just evaluating $$f(x)$$ at $$x_1$$. See Ruffini's rule.

Good moral to remember for other problems: Dividing polynomials by $$x-a$$ doesn't require division, only evaluation at $$a$$.

Simply, we have $$h(x)=\frac{f(x)}{(x-x_1)(x-x_2)}\in\mathbb{Q}[x]$$ Therefore, there is a $$q\in\mathbb{Z}$$ so that $$qh(x)\in\mathbb{Z}[x]$$ and the content of $$qh(x)$$ is $$1$$. By Gauss' Lemma, $$(x-x_1)(x-x_2)qh(x)=qf(x)$$ has content $$1$$. Since $$f\in\mathbb{Z}[x]$$, $$q$$ divides the content of $$qf(x)$$, so $$q=1$$. That is, $$h(x)\in\mathbb{Z}[x]$$.

• Some answers were downvoted, but not all; so if the downvotes were because of the question answered, then the downvotes are directed at certain users. If the downvotes were indeed because there was something wrong with the answers, it would be nice to know what was wrong. – robjohn Nov 17 at 0:53