Doubt about division between two polynomial: $x^2+10x+21$ and $x+3$

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $0$.