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