Question #c7d51

1 Answer
Dec 12, 2016

#r(x) = (2^51-1)x+2-2^51#

Explanation:

#x^51=p_(49)(x)(x^2-3x+2)+r(x)#

Here #p_(49)(x)# represents a polynomial of degree #49# and #r(x)# is the division remainder. The remainder degree is always lower than the divisor. So #p_49(x)# is the quotient, #(x^2-3x+2)# is the divisor, and #r(x)# the remainder. Note that

degree[#p_49(x)#] = #49#
degree[#(x^2-3x+2)#] = #2#
degree[#r(x)#] = #1#

then

degree[#p_(49)(x)(x^2-3x+2)#] = #51#

but #(x^2-3x+2)=(x-1)(x-2)# and

the general #1# degree polynomial representing #r(x)# is #ax+b# so

#x^51=p_(49)(x)(x-1)(x-2)+ax+b# The constants #a,b# are obtained because we know two conditions to be obeyed.

#x=1->1^51=a+b# and
#x=2->2^51=2a+b#

Solving for #a, b# we get

#r(x) = (2^51-1)x+2-2^51#