# Finding the remainder!

After spending some mind and time I got the following result:

Given any polynomial $f(x)$ with degree greater than or equal to $2$. And you know that if it is divided by $x-p$ the remainder we get is $q$. And if it is divide by $x-q$ then the remainder is $p$, then if you divide it by $(x-p)(x-q)$ the remainder you will get will be $\boxed{p+q-x}$ where $x$ is a variable and $p$ and $q$ are constants and $\boxed{p≠q}$

Proof:

Consider any $f(x)$, and let consider it's division by $x-p$ now it is necessary that degree of $f(x)$ is greater than or equal to $2$ (see why?)

$f(x)=(x-p)g_1(x)+q$

Where $g_1(x)$ is any polynomial

Similarly of we divide it by $x-q$

$f(x)=(x-q)g_2(x)+p$

Now if we divide it by $(x-p)(x-q)$

$f(x)=(x-p)(x-q)g_3(x)+r(x)$

From division algorithm of polynomial it follows that degree of $r(x)$ is smaller than $(x-p)(x-q)$ or $r(x)=0$

As degree of $(x-p)(x-q)$ will be $2$ therefore the highest possible degree of $r(x)$ is$1$. Therefore there exists $a$ and $b$ such that $r(x)=ax+b$ where it is not necessary that $a≠0$ as $r(x)$ can also be a constant in that case $a=0$ and if $r(x)=0$ then $a=0$ and $b=0$.

$f(x)=(x-p)(x-q)g_3(x)+(ax+b)$

$f(p)=(p-p)(p-q)g_3(p)+(ap+b)=ap+b$

$and$

$f(q)=(q-p)(q-q)g_3(q)+(aq+b)=aq+b$

Also as remainder of division of $f(x)$ and $(x-p)$ is $f(p)$ from remainder theorem. Therefore, $f(p)=q$ because both (i.e $f(p)$ and $q$) of them are the remainder of the same division, similarly $f(q)=p$ therefore,

$q=ap+b$ ...(1)

$and$

$p=aq+b$ ...(2)

if we subtract the (1) from (2) we get

$p-q=aq-ap$

$p-q=a(q-p)$

$a=-1$

putting it in the equation ($p=aq+b$) we get

$p=-q+b$

$b=p+q$

Therefore the remainder $r(x)=ax+b=-x+p+q=\boxed{p+q-x}$

Note:

• It is important to note that if $p=q$ then from the equation ($p-q=a(q-p)$) we can't get that $a=-1$ because in that case $p-q=q-p=0$ and as you can't divide anything by $0$, therefore $\boxed{p≠q}$ as mentioned in the statement above

Note by Zakir Husain
5 days, 2 hours ago

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