Using the remainder theorem, how do you find the remainder of #3x^5-5x^2+4x+1# when it is divided by #(x-1)(x+2)#?

1 Answer
Nov 8, 2017

#42x-39=3(14x-13).#

Explanation:

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.!