Problem no:12.
Find the remainder on dividing ($x^{100}2x^{51}+1$) by ($x^21$)?
BdMO National Secondary 2007/12

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Re: BdMO National Secondary 2007/12
$x^21=0$
$x^2=1$
$x= +1$ or, $1$
Let, $f(x)=x^{100}2x^{51}+1$
Now, the remainder(s) can be found using the Remainder theorem.
When $x=+1$
$f(x)=12(1)+1$
$f(x)=0$
When $x=1$
$f(x)=12(1)+1$
$f(x)=4$
$x^2=1$
$x= +1$ or, $1$
Let, $f(x)=x^{100}2x^{51}+1$
Now, the remainder(s) can be found using the Remainder theorem.
When $x=+1$
$f(x)=12(1)+1$
$f(x)=0$
When $x=1$
$f(x)=12(1)+1$
$f(x)=4$
Re: BdMO National Secondary 2007/12
Since the highest power of the variable in your divisor polynomial $(x^21)$ is 2, your remainder polynomial will be in the form of $ax^1+b$. Right?
Now suppose $f(x) = x^{100}  2x^{51} + 1 = p(x)\cdot(x^21) + ax + b \qquad (Eq.1)$
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 $x^21 = 0$, calculate two values of $x$ and two corresponding values of $f(x)$, and put those into $(Eq. 1)$
Share your results with us.
Now suppose $f(x) = x^{100}  2x^{51} + 1 = p(x)\cdot(x^21) + ax + b \qquad (Eq.1)$
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 $x^21 = 0$, calculate two values of $x$ and two corresponding values of $f(x)$, and put those into $(Eq. 1)$
Share your results with us.