Dividing higher-order algebraic expressions I've been self-studying from Stroud & Booth's amazing "Engineering Mathematics", and am currently on the "Algebra" section.
I've been working with division of algebraic expressions, and the book explains how to solve the simpler ones, such as:
$$(2{y}^2 - y - 10) \div (y + 2)$$
But, for the practice exercises, I have one that I absolutely have no clue where to start, and none of the examples or problem sets so far in the book come even close to it:
$$\frac{2r^3 + 5r^2 - 4r^3 + 3}{r^2 + 2r - 3}$$
Can anyone help me out, or point me in the right direction, please?
 A: Solving the equation $$2y^2-2y-10=0$$ we get
$$y^2-\frac{1}{2}y-5=0$$ so $$y_{1,2}=\frac{1}{4}\pm\sqrt{\frac{81}{4}}$$ so we get $$y_1=\frac{5}{2}$$ or $$y=-2$$ so you can write
$$\frac{(y+2)(y-\frac{5}{2})}{y+2}$$
Your second quotient can be written as
$$\frac{2r^3+5r^2-4r+3}{r^2+2r-3}=2\,r+1-3/2\, \left( r+3 \right) ^{-1}+3/2\, \left( r-1 \right) ^{-1}$$
A: $2r^3+5r^2-4r+3=2r\left(r^2+2r-3\right)+\left(r^2+2r-3\right)+6=(2r+1)(r^2+2r-3)+6$
$$\Rightarrow \frac{2r^3+5r^2-4r+3}{r^2+2r-3}=\frac{(2r+1)(r^2+2r-3)+6}{r^2+2r-3}=2r+1+\frac{6}{r^2+2r-3}$$
A: $$
\require{enclose}
\begin{array}{rl}
    2r + 1 \\[-3pt]
   r^2+2r-3 \enclose{longdiv}{2r^3+5r^2-4r+3}\kern-.2ex \\[-3pt]
      \underline{2r^3+4r^2-6r\quad\:\:} \\[-3pt]
      r^2+2r+3\, \\[-3pt]
      \underline{\: r^2+2r+3} \\[-3pt]
      \:6\,
  \end{array} \\[5pt]\begin{aligned} &\therefore 2r^3+5r^2-4r+3 \\ &= (2r+1)(r^2+2r-3) + 6 \\ &= (2r+1)(r-1)(r+3) + 6\end{aligned}
$$
or,
$$
\frac{2r^3+5r^2-4r+3}{r^2+2r-3} = \underset{\text{quotient}}{\underbrace{2r+1}} + \overset{\text{remainder}}{\overbrace{\frac{6}{r^2+2r-3}}}
$$
