When #x^4+ax^3+bx^2-2x+5# is divided by #x^2-1# the remainder is #2x+3#. What are the values of #a# and #b# ?

1 Answer
George C.
Oct 5, 2015

#a = 4# and #b = -3#

Explanation:

Use synthetic division (similar to long division of numbers) to divide #x^4+ax^3+bx^2-2x+5# by #x^2-1#. It is just like full long division of polynomials, except that we omit the powers of #x# and just work with the coefficients.

Note that the divisor (#x^2 - 1#) is represented by #1, 0, -1# not #1, -1#. We need the #0# to represent the coefficient of the missing #x# term (#x^2+0x-1#) ...

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So

#x^4+ax^3+bx^2-2x+5#

#= (x^2-1)(x^2+ax+(b+1)) + (a-2)x + (b+6)#

We are told that the remainder is #2x+3#, so

#a-2 = 2#

#b+6 = 3#

Hence #a=4# and #b=-3#

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