Question

Question #fa20c

Answer 1

See below.

Explanation:

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)`