If dividing #x^4+ax^2+bx^2-2x+5# by #x^2-1# leaves remainder #2x+3#, then what are the values of #a# and #b#?

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Answer 1 · 2015-10-03T10:42:01

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

Use synthetic division to divide #x^4+ax^3+bx^2-2x+5# by #x^2-1# (not forgetting the #0# coefficient of the #x# term in the divisor) ...

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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#