When 2x+1 is divided into #4x6^4-15x^2+px+6#, the remainder is 2, how do you find p?

1 Answer
Alan P.
Nov 14, 2015

Perform the division (using polynomial long division or synthetic division); equate the remainder to #2#, and solve for #p# to get:
#p=1#

Explanation:

#{: (,,2x^3,-x^2,-7x,+(p+7)/2,), (,,"------","------","------","--------","------"), (2x+1,")",4x^4,,-15x^2,+px,+6), (,,4x^4,+2x^3,,,), (,,"------","------",,,), (,,,-2x^3,-15x^2,,), (,,,-2x^3,x^2,,), (,,,"------","------",,), (,,,,-14x^2,+px,), (,,,,-14x^2,-7x,), (,,,,"------","------",), (,,,,,(p+7)x,+6), (,,,,,(p+7)x,+(p+7)/2), (,,,,,"--------","------"), (,,,,,,6-(p+7)/2) :}#

We are told the remainder #=2#
So
#color(white)("XXX")6-(p+7)/2 = 2#

#color(white)("XXX")(p+7)/2 = 4#

#color(white)("XXX")p+7 = 8#

#color(white)("XXX")p=1#

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