What is #(2h^4- h^3+h^2+h -3) -: (h^2-1)#?

1 Answer
Ananda Dasgupta
Feb 12, 2018

#2h^2-h+3#

Explanation:

#2h^4−h^3+h^2+h−3 #
#= 2h^4-2h^3+h^3-h^2+2h^2-2h+3h-3 #
#= 2h^3(h-1)+h^2(h-1)+2h(h-1)+3(h-1)#
# = (h-1)(2h^3+h^2+2h+3)#
# = (h-1) (2h^3+2h^2-h^2-h+3h+3)#
# = (h-1) {2h^2(h+1)-h(h+1) +3(h+1)}#
# = (h-1)(h+1)(2h^2-h+3)#

Since substituting #h=pm 1# in the expression #2h^4−h^3+h^2+h−3 # gives us zero, we know beforehand that both #h-1# and #h+1# are factors !

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