Let us denote, by #p(x)=3x^5-5x^2+4x+1,# the given
polynomial (poly.).
Noting that the divisor poly., i.e., #(x-1)(x+2),# is of degree
#2,# the degree of the remainder (poly.) sought for, must be
less than #2.#
Therefore, we suppose that, the remainder is #ax+b.#
Now, if #q(x)# is the quotient poly., then, by the Remainder Theorem,
we have, #p(x)=(x-1)(x+2)q(x)+(ax+b), or, #
# 3x^5-5x^2+4x+1=(x-1)(x+2)q(x)+(ax+b)......(star).#
#(star)" holds good "AA x in RR.#
We prefer, #x=1, and, x=-2!#
Sub.ing, #x=1# in #(star), 3-5+4+1=0+(a+b), or,#
#a+b=3...................(star_1).#
Similarly, sub.inf #x=-2# in #p(x)# gives,
#2a-b=123................(star_2).#
Solving #(star_1) and (star_2)" for "a and b,# we get,
#a=42 and b=-39.#
These give us the desired remainder,
#42x-39=3(14x-13).#
Enjoy Maths.!
