# Thread: A doubt on divisibility

1. ## A doubt on divisibility

Why is it so? (Check out the image first)

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2. ## Re: A doubt on divisibility

If $\displaystyle (x+a)$ is a factor of $\displaystyle f(x)$, then what is $\displaystyle f(-a)$ ?

3. ## Re: A doubt on divisibility

Claim:
$$(x+a)\sum_{k=0}^{2n}(-1)^k x^k a^{2n-k} = x^{2n+1}+a^{2n+1}$$

Proof:
$$(x+a)\sum_{k=0}^{2n}(-1)^k x^k a^{2n-k} = (x+a)(x^{2n}-ax^{2n-1}+a^2x^{2n-2}\pm \cdots - a^{2n-1}x+a^{2n})$$

Now, let's expand and line up terms (the top line is the summation multiplied by $x$ while the bottom line is the summation multiplied by $a$):

$$\begin{matrix} & x^{2n+1} & -ax^{2n} & + a^2x^{2n-1} & \pm \cdots & -a^{2n-1}x^2 & +a^{2n}x & \\ + & & ax^{2n} & -a^2x^{2n-1} & \pm \cdots & +a^{2n-1}x^2 & -a^{2n}x & +a^{2n+1}\end{matrix}$$

Note that the central terms all cancel out, and all that is left is $x^{2n+1}+a^{2n+1}$.

Can you do something similar for $x^{2n}-a^{2n}$?

4. ## Re: A doubt on divisibility

Originally Posted by MarkFL
If $\displaystyle (x+a)$ is a factor of $\displaystyle f(x)$, then what is $\displaystyle f(-a)$ ?
Oh yeah got it. To make f(-a) zero,we need to ... you know
Got it got it

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5. ## Re: A doubt on divisibility

Hey i have 17 questions as doubt from my text book.can i send it all to this thread

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6. ## Re: A doubt on divisibility

Originally Posted by sbjsavio
Hey i have 17 questions as doubt from my text book.can i send it all to this thread

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We would prefer separate threads for each question.

7. ## Re: A doubt on divisibility

Originally Posted by SlipEternal
We would prefer separate threads for each question.
Ok then

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8. ## Re: A doubt on divisibility

Originally Posted by sbjsavio
Ok then

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And post work on each question so we may see where you are struggling. Don't post a problem without any work/attempts shown.