A polynomial #p(x)# of degree #n>=2# has a remainder of #9# when it is divided by #(x+2)# and a remainder of #-1# when it is divided by #(x-3)#. Find the remainder of #p(x)# when it is divided by #(x^2-x-6)#. ?

1 Answer
George C.
Jun 30, 2018

#-2x+5#

Explanation:

Suppose:

#p(x) = x^2+ax+b#

Then by the remainder theorem:

#9 = p(-2) = (color(blue)(-2))^2+a(color(blue)(-2))+b \ => \ 5 = -2a+b#

#-1 = p(3) = (color(blue)(3))^2+a(color(blue)(3))+b \ => \ -10 = 3a+b#

Subtracting the first equation from the second, we find:

#-15 = 5a \ => \ a = -3#

Then:

#-10 = 3a+b = 3(-3)+b \ => \ b = -1#

So:

#p(x) = x^2 - 3x - 1 = (x^2-x-6) -2x+5#

Hence the remainder when #p(x)# is divide by #(x^2-x-6)# is #-2x+5#

Note that this linear remainder gives the required values when we put #x=-2# or #x=3# since #(x^2-x-6) = (x+2)(x-3)#. So this is the correct remainder regardless of the form of #p(x)#.

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