Q: Doubt about division between two polynomial: $x^2+10x+21$ and $x+3$ I know $(x^2+10x+21)/(x+3)$ is $x+7$
But in $(x^2+10x+21)/(x+3)$ why can't we cancel $x^2$ with $x$ and $21$ with $3$ we get $x+10x+7$ what rule is breaking here.

A: It seems that you're doing something like
$$\frac{a+b}{c+d}=\frac ac+\frac bd$$
for particular values of $a,b,c,d$.
Take $a=b=c=d$ to see that it isn't true in general.

A: Good question! I will explain my answer in two steps:
Step 1 - Proving $x+7$ is the correct answer:
We can turn division into multiplication by
\begin{align}
\frac{x^2+10x+21}{x+3}&=x+7\\
(x+3)(x+7)&=x^2+10x+21
\end{align}
We can expand the brackets
\begin{align}
&\quad(x+3)(x+7)\\
&=x\cdot x+x\cdot7+3\cdot x+3\cdot7\\
&=x^2+7x+3x+21\\
&=x^2+10x+21
\end{align}
Therefore, $x+7$ is the correct answer.
Step 2 - Proving $x+10x+7=11x+7$ is not a correct answer:
We can easily substitute this in, expand the brackets and get the wrong answer. To make this clearer, I have made an example where $x=10$. What you did was:
\begin{align}
&\quad\frac{10^2+10\cdot10+21}{1\cdot10+3}\\
&=\frac{10^2}{10}+10\cdot10+\frac{21}{3}
\end{align}
Which is obviously wrong.

A: You are interpreting letters and polynomial as numbers. But you can see them as function (in this case $x$).
So, you have to use polynomial division rule. There are two type of it: traditional division and Ruffini's law. Here you can use both and obtain:
$$(x^2+10x+21):(x+3)=x+7$$
And the remainder is $0$.
