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P(x)=Ax^3+Bx^2+Cx+D, where A, B, C, D are constants.

P(1)=0: 0=A+B+C+D

P(-1)=4=-A+B-C+D, so 2B+2D=4 and B+D=2, D=2-B; also, 2A+2C=-4 and A+C=-2, C=-2-A

P(2)=28=8A+4B+2C+D

P(-2)=0=-8A+4B-2C+D, so 8B+2D=28, 4B+D=14; also, 16A+4C=28, 4A+C=7.

Substitute D=2-B: 4B+2-B=14, 3B=12, B=4, D=-2

Substitute C=-2-A: 4A-2-A=7, 3A=9, A=3, C=-5.

P(x)=3x^3+4x^2-5x-2.

Check: P(1)=0; P(-1)=-3+4+5-2=4; P(2)=24+16-10-2=28; P(-2)=-24+16+10-2=0. All OK!

by Top Rated User (775k points)

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