How do you use synthetic division to divide #x^4-4x^2+7x+15# divided by x+4?

1 Answer
Jan 29, 2016

Answer:

#x^3-4x^2+12x-41# with a remainder of 179.

Explanation:

#x^4-4x^2+7x+15# divided by #x+4# gives #x^3# together with a remainder, because the high order term of the divisor, namely x, times #x^3# yields the high order term of the polynomial being divided, namely #x^4#. To find the remainder, multiply #x+4# times #x^3# and get #x^4+4x^3#. Subtract that from #x^4-4x^2+7x+15# to get #-4x^3-4x^2+7x+15#. That is the initial remainder.

Divide that again by x+4, which goes #-4x^2# times, because the high order term of the divisor, namely x, times #-4x^2#, yields the high order term of this initial remainder, namely, #-4x^3#. To find the new remainder, multiply #x+4# times #-4x^2# and get #-4x^3-16x^2#. Subtract that from the initial remainder, namely #-4x^3-4x^2+7x+15# and get the secondary remainder, namely #12x^2+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 #12x^2+48#. Subtract this from the secondary remainder, namely #12x^2+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:
#x^3#, #-4x^2#, #12x#, and #-41#. Adding these together yields the polynomial #x^3-4x^2+12x-41# together with the final remainder #179#.