Divide f cubed plus 64 by f plus 4.

Step-by-step solution

1. 1

   Write out every coefficient, including the missing ones. $f^3+64$ has no $f^2$ or $f$ term, so the coefficient list is $1,\,0,\,0,\,64$. Forgetting the placeholder zeros is the single most common error in synthetic division.

2. 2

   Identify the synthetic-division root. The divisor $f+4$ is zero at $f=-4$, so $-4$ goes in the corner box - the sign is opposite to the one written in the divisor.

3. 3

   Run the algorithm. Bring down $1$; $1\times(-4)=-4$, and $0+(-4)=-4$; $-4\times(-4)=16$, and $0+16=16$; $16\times(-4)=-64$, and $64+(-64)=0$. The row reads $1,\ -4,\ 16,\ 0$.

4. 4

   Read the quotient and remainder. The last entry $0$ is the remainder, and the first three give a degree-2 quotient: $\frac{f^3+64}{f+4} = f^2-4f+16.$

5. 5

   Recognise why the remainder is zero. $64=4^3$, so this is a sum of cubes: $a^3+b^3=(a+b)(a^2-ab+b^2)$ with $a=f$, $b=4$ gives exactly $(f+4)(f^2-4f+16)$. By the factor theorem, $(-4)^3+64=0$ guaranteed a zero remainder.

6. 6

   Check by multiplying back. $(f+4)(f^2-4f+16) = f^3-4f^2+16f+4f^2-16f+64 = f^3+64$. Every middle term cancels, as it must.