5
$\begingroup$

Question: Show that the polynomial $f(x)=(x+1)^{2n} +(x+2)^n - 1$ is divisible by $g(x) = x^2+3x+2$, where $n$ is an integer.

I have tried to use mathematical induction. The basis case wasn't that difficult, but when it comes to the inductive step itself, I got a bit confused.

Is it possible to prove this by mathematical induction, and is binominal expansion required at that step?

$\endgroup$
  • 1
    $\begingroup$ Artem, For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – The Chaz 2.0 Nov 3 '12 at 18:58
9
$\begingroup$

Putting the zeros of $x^2+3x+2=0$ i.e., $-1,-2$ one by one, in $f(x)=(x+1)^{2n}+(x+2)^n-1$

we get, $f(-1)=(-1+1)^{2n}+(-1+2)^n-1=0$

and $f(-2)=(-2+1)^{2n}+(-2+2)^n-1=0$

(i)So, using Remainder Theorem,

$(x+1)\mid f(x)$ and $(x+2)\mid f(x)\implies lcm(x+1,x+2)\mid f(x)$

(ii) Alternatively,

$\frac{(x+1)^{2n}+(x+2)^n-1}{x+1}=(x+1)^{2n-1}+\frac{(x+2)^n-1}{x+1}$

Now $x+1(=x+2-1)\mid \{(x+2)^{2n-1}-1\} $ as $(a-b)\mid (a^n-b^n)--->(1)$

So, $(x+1)\mid f(x).$

$\frac{(x+1)^{2n}+(x+2)^n-1}{x+2}=\frac{\{(x+1)^2\}^n-1}{x+2}+(x+2)^{n-1}$

Using $(1),\{(x+1)^2\}^n-1$ is divisible by $(x+1)^2-1=x^2+2x=x(x+2)$

So, $(x+2)\mid f(x).$

$lcm(x+1,x+2)=(x+1)(x+2)=x^2+3x+2$ (prove this)

$\endgroup$
  • 1
    $\begingroup$ Very nicely put... $\endgroup$ – SyndicatorBBB Nov 3 '12 at 18:56
  • 1
    $\begingroup$ @Guy, welcome anytime. $\endgroup$ – lab bhattacharjee Nov 3 '12 at 18:58
  • $\begingroup$ Thank you @lab bhattacharjee! $\endgroup$ – Artem Nov 3 '12 at 21:38
  • $\begingroup$ @Artem, my pleasure. $\endgroup$ – lab bhattacharjee Nov 4 '12 at 7:49
4
$\begingroup$

Hint $\rm\: mod\ g\!:\,\ e_i^2\! = e_i\Rightarrow\: e_1^n + e_2^n =\, e_1\! + e_2.\ $ Google (orthogonal) idempotents to learn more.

$\endgroup$
2
$\begingroup$

Warning: Non-Induction Proof Put $z = x+1$ and use $z$ rather than $x$ but everything else is the same. So let's define $$f_n(z)=z^{2n} + (z+1)^n - 1.$$ Obviously $z|f_n(z)$ and $z$ is coprime to $z+1$ so we just need to show that $z+1|f_n(z)$ but that is equivalent to $-1$ being a root of the polynomial, which we verify it is by evaluation.

$\endgroup$
1
$\begingroup$

I don't have an answer for you because I didn't exercise this stuff a long time but I did find a reference that might assist you.

Please try : http://www.enotes.com/math/q-and-a/determine-m-polynomial-p-divisible-by-x-3-p-x-3-mx-188671/

Using the remainder theorem "http://en.wikipedia.org/wiki/Polynomial_remainder_theorem"

Hope It helps a little bit,

Guy

$\endgroup$

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.