BdMO National Secondary 2007/12

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samiul_samin
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BdMO National Secondary 2007/12

Unread post by samiul_samin » Wed Feb 21, 2018 11:46 pm

Problem no:12.
Find the remainder on dividing ($x^{100}-2x^{51}+1$) by ($x^2-1$)?

Ragib Farhat Hasan
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Re: BdMO National Secondary 2007/12

Unread post by Ragib Farhat Hasan » Fri Oct 19, 2018 2:04 am

$x^2-1=0$
$x^2=1$
$x= +1$ or, $-1$

Let, $f(x)=x^{100}-2x^{51}+1$
Now, the remainder(s) can be found using the Remainder theorem.

When $x=+1$
$f(x)=1-2(1)+1$
$f(x)=0$

When $x=-1$
$f(x)=1-2(-1)+1$
$f(x)=4$

SMMamun
Posts: 57
Joined: Thu Jan 20, 2011 6:57 pm

Re: BdMO National Secondary 2007/12

Unread post by SMMamun » Fri Oct 26, 2018 3:47 pm

Since the highest power of the variable in your divisor polynomial $(x^2-1)$ is 2, your remainder polynomial will be in the form of $ax^1+b$. Right?

Now suppose $f(x) = x^{100} - 2x^{51} + 1 = p(x)\cdot(x^2-1) + ax + b \qquad (Eq.1)$

You don't know about the quotient polynomial $p(x)$, and you don't need to know it either -- what you need is eliminate $p(x)$. But how can you eliminate the involvement of $p(x)$ and find two simultaneous equations with $a$ and $b$? Very easy: make $x^2-1 = 0$, calculate two values of $x$ and two corresponding values of $f(x)$, and put those into $(Eq. 1)$

Share your results with us. :D

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