[tex]4x^3=2x\cdot2x^2[/tex], and [tex]2x(2x^2+3x)=4x^3+6x^2[/tex]. Subtract this from the numerator to get a remainder of
[tex](4x^3+10x^2+12x+1)-(4x^3+6x^2)=4x^2+12x+1[/tex]
[tex]4x^2=2\cdot2x^2[/tex], and [tex]2(2x^2+3x)=4x^2+6x[/tex]. Subtract this from the previous remainder to get a new remainder of
[tex](4x^2+12x+1)-(4x^2+6x)=6x+1[/tex]
[tex]6x[/tex] is not divisible by [tex]2x^2[/tex], so we're done, and we've found that
[tex]\dfrac{4x^3+10x^2+12x+1}{2x^2+3x}=2x+\dfrac{4x^2+12x+1}{2x^2+3x}[/tex]
[tex]\dfrac{4x^3+10x^2+12x+1}{2x^2+3x}=2x+2+\dfrac{6x+1}{2x^2+3x}[/tex]
so that the quotient is [tex]2x+2[/tex] with remainder [tex]6x+1[/tex].
