Divide x^3 by x + 1 and write the answer as a quotient plus a remainder term.

Step-by-step solution

1. 1

   Write the numerator with every power present. Missing terms must be held open with zero coefficients or the columns will misalign:

   $x^3+0x^2+0x+0$

2. 2

   First division step. $\tfrac{x^3}{x}=x^2$; multiply back and subtract:

   $x^2(x+1)=x^3+x^2\quad\Longrightarrow\quad\left(x^3+0x^2\right)-\left(x^3+x^2\right)=-x^2$

3. 3

   Second step. $\tfrac{-x^2}{x}=-x$; multiply and subtract:

   $-x(x+1)=-x^2-x\quad\Longrightarrow\quad\left(-x^2+0x\right)-\left(-x^2-x\right)=x$

4. 4

   Third step. $\tfrac{x}{x}=1$; multiply and subtract:

   $1(x+1)=x+1\quad\Longrightarrow\quad(x+0)-(x+1)=-1$

   The degree of $-1$ is below the degree of $x+1$, so the division stops with quotient $x^2-x+1$ and remainder $-1$.

5. 5

   Assemble the answer and confirm with the remainder theorem.

   $\frac{x^3}{x+1}=x^2-x+1-\frac{1}{x+1},\qquad x\neq -1$

   The remainder theorem predicts the remainder as $f(-1)=(-1)^3=-1$ $\checkmark$. Equivalently $(x+1)(x^2-x+1)=x^3+1$, so $x^3=(x+1)(x^2-x+1)-1$ $\checkmark$.