Question #fa20c

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Answer 1 · 2018-02-24T23:50:59

See below.

The identities

#I_1->a (x - 1)^2 + b (2 x - 1) (x - 1) + c (x + 3) equiv (5 - x)#

#I_2->a (x - 2)^2 + b (x - 1)^2 + c equiv (3 x^2 - 8)#

are obtained after expanding and grouping coefficients given

#I_1->{(a+b+3c=5),(2a+3b=1),(a+2b=0):}#

and solving we obtain #a=4,b=-2,c=1#

so

#I_1->4 (x - 1)^2 -2 (2 x - 1) (x - 1) + (x + 3) equiv (5 - x)#

analogously we have

#I_2->-3 (x - 2)^2 +6 (x - 1)^2 -2 equiv (3 x^2 - 8)#