If you're familiar with synthetic division, but not the extended form (which allows you easily compute the quotient/remainder when dividing a polynomial by another polynomial of degree greater than 1), then you can perform two steps of SD.
Instead of dividing by [tex]x^2+3x[/tex], first divide by [tex]x[/tex], then by [tex]x+3[/tex] (since [tex]x^2+3x=x(x+3)[/tex]). So we have
0 | 1 4 2 1 4
... | 0 0 0 0
= = = = = = = = = = = =
... | 1 4 2 1 4
which translates to
[tex]\dfrac{x^4+4x^3+2x^2+x+4}x=x^3+4x^2+2x+1+\dfrac4x[/tex]
Ignoring the remainder term for now, the next round of SD yields
-3 | 1 4 2 1
... | -3 -3 3
= = = = = = = = = =
... | 1 1 -1 4
which translates to
[tex]\dfrac{x^3+4x^2+2x+1}{x+3}=x^2+x-1+\dfrac4{x+3}[/tex]
Now, putting everything together, we have
[tex]\dfrac{x^4+4x^3+2x^2+x+4}{x^2+3x}=\dfrac{x^3+4x^2+2x+1+\frac4x}{x+3}[/tex]
[tex]=x^2+x-1+\dfrac4{x+3}+\dfrac4{x(x+3)}[/tex]
[tex]=x^2+x-1+\dfrac{4x+4}{x^2+3x}[/tex]
which is to say the remainder upon dividing [tex]x^4+4x^3+2x^2+x+4[/tex] by [tex]x^2+3x[/tex] is [tex]4x+4[/tex].
