Remainder Theorem

**acc100jt** · July 26th 2009, 07:16 PM

Find the remainder when $x^{100}+5$ is divided by $x^2-2x+1$

Anyone can help? Thanks

**VonNemo19** · July 26th 2009, 07:37 PM

Originally Posted by acc100jt

Find the remainder when $x^{100}+5$ is divided by $x^2-2x+1$

Anyone can help? Thanks

$x^2-2x+1=(x-1)(x-1)$

The remainder theorem states that if a polynomial $f(x)$ is divided by $x-k$ , the remainder is $r=f(k)$ .

So, if I were to divide $x^{100}+5$ by $(x-1)$ , this implies that the remainder is $(1)^{100}+5=6$ . If I were to do it again...

I think. It's been a long time man.

**acc100jt** · July 26th 2009, 07:43 PM

i understand the first part. but what do u mean "do it again"?what is the dividend if i wan to divide again?

**VonNemo19** · July 26th 2009, 08:05 PM

Originally Posted by acc100jt

i understand the first part. but what do u mean "do it again"?what is the dividend if i wan to divide again?

By the Division Algorithm

$f(x)=d(x)q(x)+r(x)$

The way to use this is

$\frac{f(x)-r(x)}{d(x)}=q(x)$

where $f(x)=x^{100}+5,r(x)=6,\text{ and },q(x)=\text{ the new function to be divided}$

**acc100jt** · July 26th 2009, 08:05 PM

this is what I've tried,

Let $f(x)=x^{100}+5$

$f(1)=6$

So
$f(x)=(x-1)Q(x)+6$ ....(1)
$f(x)-6=(x-1)Q(x)$
$x^{100}-1=(x-1)Q(x)$
$(x-1)(x^{99}+x^{98}+...+x+1)=(x-1)Q(x)$

Thus $Q(x)=x^{99}+x^{98}+...+x+1$

Since $Q(1)=100$
So $Q(x)=(x-1)P(x)+100$ , substitute this into (1)

$f(x)=(x-1)[(x-1)P(x)+100]+6$
$f(x)=(x-1)^{2}P(x)+100x-94$

Hence the remainder is $100x-94$

Am I correct?

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