How do you divide $\dfrac{{{x^2} + 12x + 16}}{{x + 4}}$?

Hint: First we will reduce the equation further if possible. Then we will try to factorise the terms in the equation. Then solve the equation by using the quadratic formula and finally evaluate the value of the variable accordingly.

Complete step-by-step answer:
We will start off by mentioning all the steps and then the calculations.
So, the first term of dividend polynomial that is the one inside the division box divided by the first term of the divisor which is outside the box, this will be the first term of the quotient.
Now next we will multiply the divisor polynomial by the first term of the quotient to get another polynomial and then we will subtract this polynomial from the dividend. Finally, bring down the next term in the dividend. Now repeat all these above-mentioned steps until you reach the final remainder.
At some point you will reach a number that you will not have any variable. That term will be our remainder.
\[
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x + 8 \\
  x + 4|\overline {{x^2} + 12x + 16} \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\underline {{x^2} + 4x} \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,8x + 16 \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline {8x + 32} \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 16 \\
 \]
Hence, the remainder of the expression $\dfrac{{{x^2} + 12x + 16}}{{x + 4}}$ is $ - 16$.
So, the correct answer is “$ - 16$”.

Note: Long division with polynomials is hard to explain in words. Long division is a standard division algorithm suitable for dividing multi-digit numbers that is simple enough to perform by hand. It breaks down a division problem into a series of easier steps.
Do not solve all the equations simultaneously. Solve all the equations separately, so that you don’t miss any term of the solution. Check if the solution satisfies the original equation completely. If any term of the solution doesn’t satisfy the equation, then that term will not be considered as a part of the solution.