x^4-4x^2+7x+15x4−4x2+7x+15 divided by x+4x+4 gives x^3x3 together with a remainder, because the high order term of the divisor, namely x, times x^3x3 yields the high order term of the polynomial being divided, namely x^4x4. To find the remainder, multiply x+4x+4 times x^3x3 and get x^4+4x^3x4+4x3. Subtract that from x^4-4x^2+7x+15x4−4x2+7x+15 to get -4x^3-4x^2+7x+15−4x3−4x2+7x+15. That is the initial remainder.
Divide that again by x+4, which goes -4x^2−4x2 times, because the high order term of the divisor, namely x, times -4x^2−4x2, yields the high order term of this initial remainder, namely, -4x^3−4x3. To find the new remainder, multiply x+4x+4 times -4x^2−4x2 and get -4x^3-16x^2−4x3−16x2. Subtract that from the initial remainder, namely -4x^3-4x^2+7x+15−4x3−4x2+7x+15 and get the secondary remainder, namely 12x^2+7x+1512x2+7x+15
Divide the secondary remainder by the divisor x+4x+4 which goes 12x12x times. To get the third remainder multiply 12x12x by x+4x+4 which yields 12x^2+4812x2+48. Subtract this from the secondary remainder, namely 12x^2+7x+1512x2+7x+15, to get the third remainder, namely 41x+1541x+15.
Divide the third remainder by x+4x+4 which goes 41 times. To get the final remainder multiply 4141 times x+4x+4 which yields 41x+16441x+164. Subtracting that from the third remainder,
namely 41x+1541x+15 yields 179. The successive divisors are:
x^3x3, -4x^2−4x2, 12x12x, and -41−41. Adding these together yields the polynomial x^3-4x^2+12x-41x3−4x2+12x−41 together with the final remainder 179179.
