I have the formula: $$ \frac{x^3 + 20x^2 + 100x}{x^3+31x^2+320x+1100} $$
How can I simplify them? I have no idea how to get started. I can imagine that you can look at it as two functions and then work your way along.
$f(x) = x^3 + 20x^2 + 100x$
$g(x) = x^3+31x^2+320x+1100$
And then you try something like this:
$$ \frac{f(x)}{g(x)} $$
But I am not sure and also do not know how to start here
On
When we use long division to calculate $\frac{a}{b} \in \mathbb{Q}$ what we're doing is looking for integers $q$ and an $r$ such that $r < a$ and $a=qb+r$. Polynomials follow a similiar idea except that now for two polynomials $a$ and $b$ to calculate $\frac{a}{b}$ we want polynomials $q$ and $r$ so that $a =qb + r$ with $\deg r < \deg g$. This is basically the same idea except we're using the degree of the polynomial to determine when to stop rather than the usual remainder.
In your case since $\deg a = \deg b$ it can be simplified. As with regular long divison we need to choose a suitable multiple of $b$ to subtract off of $a$. What's nice is that since we're only interested in the degree of the polynomial is that this is much easier. Say the leading term of $a$ is $wx^j$ and the leading term and $ux^k$ with $k < j$ and $u,w \neq 0$. Then since we have $\frac{w}{u}x^{j-k}ux^{k}=wx^j$ so $\frac{w}{u}x^{j-k}$ is always the correct multiple. Since we only have to consider the leading terms one at a time lets look at a couple simple examples.
Say we have $\frac{x^2}{x^2+1}$ here $\deg a = \deg b$ so we can simplify and since $w=u=1$ we only have to consider the powers. Since the degrees at the same we subtract just one copy of $x^2+1$ from $x^2$ to get $x^2-(x^2+1)=-1$ and since $\deg -1 < \deg x^2+1$ we can stop. Notice that $x^2 = 1(x^2+1) + (-1)$ is in the form $a=bq+r$ mentioned earlier.
Now say you're given $\frac{x^4+1}{x}$. We notice again that $u=w=1$ and we need to multiply $x$ by $x^3$ to get an appropriate multiple to lower the degree of the numerator. So we have $x^4+1-x^4=1$ just like we did before. We can see the difference when we write the $a=bq+r$ form however since now $x^4+1 = x(x^3)+1$. Writing it in this form gives you an easy way to double check your work too.
Notice that despite the degrees of both polynomials being rather different in both examples it was just one step to get to the remainder and that's because there were no intermediate terms. If you work on $\frac{x^4+x^3+x^2+x+1}{x-2}$ there will be more steps for each coefficient that doesn't cancel out when you subtract. The coefficients are also a lot easier to deal with when the polynomial is normalized to have a leading coefficient to be $1$ but it's not required.
Can you take it from here?
There is a method, same as Euclidean algorithm for finding gcd of integers. This finds the gcd of the polynomials in $\mathbb Q[x]$ meaning I allow rational coefficients. You can learn this!
ummm: instead of "back-substituion" I like to write the business as a continued fraction.
$$ \left( x^{3} + 20 x^{2} + 100 x \right) $$
$$ \left( x^{3} + 31 x^{2} + 320 x + 1100 \right) $$
$$ \left( x^{3} + 20 x^{2} + 100 x \right) = \left( x^{3} + 31 x^{2} + 320 x + 1100 \right) \cdot \color{magenta}{ \left( 1 \right) } + \left( - 11 x^{2} - 220 x - 1100 \right) $$ $$ \left( x^{3} + 31 x^{2} + 320 x + 1100 \right) = \left( - 11 x^{2} - 220 x - 1100 \right) \cdot \color{magenta}{ \left( \frac{ - x - 11 }{ 11 } \right) } + \left( 0 \right) $$
Here I begin the "continued fraction" part. It begins with two fractions, the legitimate $0/1$ and the fake $1/0.$
$$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( 1 \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ - x - 11 }{ 11 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - x }{ 11 } \right) }{ \left( \frac{ - x - 11 }{ 11 } \right) } $$ $$ \left( x \right) \left( \frac{ 1}{11 } \right) - \left( x + 11 \right) \left( \frac{ 1}{11 } \right) = \left( -1 \right) $$ $$ \mbox{GCD} = \color{blue}{ \left( x^{2} + 20 x + 100 \right) } $$ $$ \left( x^{3} + 20 x^{2} + 100 x \right) = \left( x \right) \cdot \color{blue}{ \left( x^{2} + 20 x + 100 \right) } + \left( 0 \right) $$ $$ \left( x^{3} + 31 x^{2} + 320 x + 1100 \right) = \left( x + 11 \right) \cdot \color{blue}{ \left( x^{2} + 20 x + 100 \right) } + \left( 0 \right) $$ $$ \mbox{GCD} = \color{blue}{ \left( x^{2} + 20 x + 100 \right) } $$ $$ \left( x^{3} + 20 x^{2} + 100 x \right) \left( \frac{ 1}{11 } \right) - \left( x^{3} + 31 x^{2} + 320 x + 1100 \right) \left( \frac{ 1}{11 } \right) = \left( - x^{2} - 20 x - 100 \right) $$