How do you divide #(4x^4 -12x^3+5x+3)/((x^2 + 4x + 4) )#?

I like to long divide the coefficients, not forgetting to include #0#'s for any missing powers of #x#. In our example that means the #x^2# term of the dividend.

The process is similar to long division of numbers.

Write the dividend #4, -12, 0, 5, 3# under the bar and the divisor #1, 4, 4# to the left of the bar.

Write the first term #color(blue)(4)# of the quotient above the bar, choosing it so that when multiplied by the divisor it matches the leading term of the dividend.

Write the product #4, 16, 16# of this first term and the divisor under the dividend and subtract it. Bring down the next term #5# of the dividend alongside it to give the running remainder.

Write the second term #color(blue)(-28)# of the quotient above the bar, choosing it so that when multiplied by the divisor it matches the leading term of our running remainder.

Write the product #-28, -112, -112# of this second term and the divisor below the remainder and subtract it. Bring down the next term #3# of the dividend alongside it to give the running remainder.

Write the third term #color(blue)(96)# of the quotient above the bar, choosing it so that when multiplied by the divisor it matches the leading term of the running remainder.

Write the product #96, 384, 384# of this third term and the divisor under the running remainder and subtract it to give the remainder #-277, -381#.

This remainder is shorter than the divisor and there are no more terms to bring down from the dividend, so this is our final remainder and where we stop.

We find that:

#(4x^4-12x^3+5x+3)/(x^2+4x+4) = 4x^2-28x+96 + (-277x-381)/(x^2+4x+4)#

Or if you prefer:

#4x^4-12x^3+5x+3#

#= (x^2+4x+4)(4x^2-28x+96) + (-277x-381)#