Divide j cubed minus 64 by j minus 4.

Step-by-step solution

1. 1

   List the coefficients with placeholders. $j^3-64$ becomes $1,\,0,\,0,\,-64$; the absent $j^2$ and $j$ terms must appear as zeros or the columns will shift.

2. 2

   Choose the root. $j-4=0$ gives $j=4$, so the corner value is $+4$ - the opposite sign to the $-4$ written in the divisor.

3. 3

   Carry out the synthetic division. Bring down $1$; $1\times4=4$ and $0+4=4$; $4\times4=16$ and $0+16=16$; $16\times4=64$ and $-64+64=0$. The bottom row is $1,\ 4,\ 16,\ 0$.

4. 4

   Read off the answer. $\frac{j^3-64}{j-4} = j^2+4j+16, \qquad \text{remainder } 0.$

5. 5

   Compare with the sum-of-cubes case. The identity $a^3-b^3=(a-b)(a^2+ab+b^2)$ produces all plus signs in the quadratic factor, whereas $a^3+b^3=(a+b)(a^2-ab+b^2)$ alternates. Mixing the two up is the classic slip; synthetic division settles it mechanically.

6. 6

   Verify. $(j-4)(j^2+4j+16) = j^3+4j^2+16j-4j^2-16j-64 = j^3-64$, confirming the quotient.