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everyone

I have this problem

If a and b are integers, we say that $b$ divides $a$, and we write $b | a$ , if there is an integer c such that $a = bc$. Consider the polynomial f given by: $$f (t) = t ^ {n} + a_ {n-1} t ^ {n-1} + ... + a_ {0}$$ where the coefficients $a_ {n-1}, ..., a_ {0}$ are integers. Prove that if an integer a is a root of f, then a divides $a_ {0}$

I do this:

If a is a root and $a_ {0}=0$ the proof is immediate because $a_ {0}=c$*a* then c=0, now if a is a root I can say that $$f (t) = (x-a)q$$ with q a polynomial with low degree that f, if I do the multiplication of $(x-a)q = (x-a)(q ^ {n-1} +q_ {n-2} q ^ {n-2} + ... + q_ {0})$ at the last term $(x-a)(q_ {0})=xq_ {0}-aq_ {0}$ where

$-aq_ {0}=a_{0}$ $\Rightarrow{}$ $c=q_ {0}$ because $-ac=a_{0}$

I understand the idea, but my proof isn't formal

Any idea??

Thanks for your help.

Git Gud
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user63192
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5 Answers5

4

If $a$ is the a root of $f(t)$, we have $$a^n + a_{n-1}a^{n-1} + a_{n-2} a^{n-2} + \cdots + a_1 a + a_0 = 0 $$ This gives us $$a \left(a^{n-1} + a_{n-1} a^{n-2} + \cdots + a_1\right) = -a_0$$ Hence, $$a \vert a_0$$

3

If $a$ is a root of $f(t)$, then $$f(a) = a_na^n + a_{n-1}a^{n-1} + \cdots + a_1a+a_0 = 0.$$

Move $a_0$ to the right-hand side:

$$a_na^n + \cdots a_1 a = -a_0$$

Factor out $a$ on the left-hand side:

$$a(a_na^{n-1}+a_{n-1}a^{n-2}+\cdots+a_1) = -a_0.$$

If $a_j$ are all integers, what can be said about the term in parentheses?

Emily
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  • what can be said about the term in parentheses?

    that ac=-a0 and c is the term in parentheses and c obviously is interger

    Thank you very much :D Nice day

     – user63192 Mar 07 '13 at 23:43
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If $a$ is a root then $a^n+a_{n-1}a^{n-1}+\cdots+a_1a+a_0=0$. Thus $$a_0=a(-1)\left(a^{n-1}+a_{n-1}a^{n-2}+\cdots+a_1\right).$$

André Nicolas
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Why so much complication...

If $a$ is a root of $t^n + a_{n - 1} t^{n - 1} + \ldots a_0$, then: $$ a^n + a_{n - 1} t^{n - 1} + \ldots + a_0 = 0 $$ The right hand side is divisible by $a$, and then so is the left hand side. Each term of the left hand side, except possibly the last one, is divisible by $a$ (it has positive powers of $a$), and so $a_0$ must also be divisible by $a$.

vonbrand
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Since none of the answers addressed your proof, I would like to emphasize that it can be made formal, and it leads to a nice proof. Namely, by the Factor Theorem, since $\rm\:f(a) = 0,\:$ we can write $\rm\:f(x) = (x-a)\,g(x),\:$ for $\rm\:g(x)\in \Bbb Z[x].\:$ Thus $\rm\:f(0) = -a\,g(0),\ $ so $\rm\ a\mid f(0)= a_0.\ \ $ QED

Remark $\ $ This approach is conceptually superior to rearranging the polynomial since it generalizes to higher degree (algebraic) integers using a minimal polynomial in place of $\rm\:x-a.$

Math Gems
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